Prepare a very detailed Logarithmic equations concept clarity efficient video at jee advanced level. It's should be atleast 5 minute long. Use this for reference ---LOGARITHMIC EQUATIONS While solving logarithmic equations, we tend to simplify the equation. Solving equation after simplification may give some roots which may not define all the terms in the initial equation. Thus, while solving equations involving logarithmic function, we must take care of domain of the equation. ILLUSTRATION 1.43 Solve log_4 8 + log_4 (x + 3) – log_4 (x – 1) = 2. Sol. log_4 8 + log_4 (x + 3) – log_4 (x – 1) = 2 or log_4 (8(x + 3) / (x – 1)) = 2 or 8(x + 3) / (x – 1) = 4^2 or 8x + 24 = 16x – 16 or x + 3 = 2x – 2 => x = 5 Also for x = 5 all terms of the equation are defined. ILLUSTRATION 1.44 Solve log (–x) = 2 log (x + 1). Sol. By definition, x < 0 and x + 1 > 0. => –1 < x < 0 Now log (–x) = 2 log (x + 1) or –x = (x + 1)^2 or –x = x^2 + 2x + 1 or x^2 + 3x + 1 = 0 or x = (–3 ± sqrt(3^2 – 4(1)(1))) / 2 x = (–3 ± sqrt(9 – 4)) / 2 x = (–3 ± sqrt(5)) / 2 Hence, x = (–3 + sqrt(5)) / 2 is the only solution. (sqrt(5) approx 2.23, so (-3+2.23)/2 approx -0.385 which is between -1 and 0. (-3-sqrt(5))/2 approx (-3-2.23)/2 approx -2.615 which is less than -1). Thus (-3 - sqrt(5)) / 2 < -1 ILLUSTRATION 1.45 Solve log_2 (3x – 2) = log_1/2 x. Sol. log_2 (3x – 2) = log_1/2 x = – log_2 x = log_2 x^(-1) or 3x – 2 = x^(-1) or 3x^2 – 2x = 1 => 3x^2 – 2x – 1 = 0 or (3x + 1)(x – 1) = 0 or x = 1 or x = –1/3. But log_2 (3x – 2) and log_1/2 x are meaningful if x > 2/3. Hence, x = 1. ILLUSTRATION 1.46 Solve 2^(x+2) * 2^(x/(x-1)) = 9. Sol. Taking log of both sides, we get log [ 2^(x+2) * 2^(x/(x-1)) ] = log 9 (x + 2) log 2 + (x / (x – 1)) log 2 = log 9 or (x + 2) log 2 + (x / (x – 1)) log 2 = log 3^2 or (x + 2 + x / (x – 1)) log 2 = 2 log 3 or ( (x+2)(x-1) + x ) / (x-1) log 2 = 2 log 3 or ( x^2 + x – 2 + x ) / (x – 1) log 2 = 2 log 3 or (x^2 + 2x – 2) / (x – 1) = 2 log 3 / log 2 or (x^2 + 2x – 2) / (x – 1) = 2 log_2 3 or x^2 + 2x – 2 = 2(x – 1) log_2 3 or x^2 + 2x – 2 = (2x – 2) log_2 3 or x^2 + 2x – 2 – (2x – 2) log_2 3 = 0 or x^2 + 2x(1 – log_2 3) – 2(1 – log_2 3) = 0 Let's look at the right column derivation provided in the image which follows a different approach from the equation with logs: Starting from: (x+2) log 2 + x/(x-1) log 27 = log 9 This equation is likely derived by taking log base 3 of the original equation 2^(x+2) * 2^(x/(x-1)) = 9. log_3 [ 2^(x+2) * 2^(x/(x-1)) ] = log_3 9 (x+2) log_3 2 + x/(x-1) log_3 2 = 2 This does not match the log 27 term. Let's re-read the original equation again. It is 2^(x+2) * 2^(x/(x-1)) = 9. Let's assume the equation in the solution is correct and try to follow it. (x+2) log 2 + x/(x-1) log 27 = log 9 or (x+2) log 2 + x/(x-1) 3 log 3 = 2 log 3 Divide by log 3 (assuming log base 10 or e): (x+2) log_3 2 + 3x/(x-1) = 2 or (x+2) log_3 2 = 2 - 3x/(x-1) or (x+2) log_3 2 = (2(x-1) - 3x) / (x-1) or (x+2) log_3 2 = (2x - 2 - 3x) / (x-1) or (x+2) log_3 2 = (-x - 2) / (x-1) or (x+2) log_3 2 + (x+2) / (x-1) = 0 or (x+2) [ log_3 2 + 1 / (x-1) ] = 0 => x+2 = 0 or log_3 2 + 1/(x-1) = 0 => x = -2 or 1/(x-1) = -log_3 2 => x = -2 or x-1 = -1/log_3 2 = -log_2 3 => x = -2 or x = 1 - log_2 3 Let's check the solution line by line from the image: (x+2) log 2 + x/(x-1) log 27 = log 9 or (x+2) log 2 + x/(x-1) 3 log 3 = 2 log 3 or (x + 2) log 2 + (3x / (x-1)) log 3 = 2 log 3 This third line in the solution seems to have a typo. It should be (x + 2) log 2 + (3x/(x-1)) log 3 = 2 log 3. Let's assume they are using base 3 logs as shown previously. (x+2) log_3 2 + x/(x-1) log_3 27 = log_3 9 (x+2) log_3 2 + x/(x-1) * 3 = 2 (x+2) log_3 2 + 3x/(x-1) = 2 (x+2) log_3 2 = 2 - 3x/(x-1) = (2x - 2 - 3x) / (x-1) = (-x-2) / (x-1) (x+2) log_3 2 + (x+2) / (x-1) = 0 (x+2) [log_3 2 + 1/(x-1)] = 0 x = -2 or 1/(x-1) = -log_3 2 = log_3 (1/2) x-1 = 1 / log_3(1/2) = log_(1/2) 3 x = 1 + log_(1/2) 3 Let's look at the provided result lines again: (x+2) [ log_2 + log 3 / (x-1) ] = 0 -- This seems incorrect based on previous steps. Let's re-examine the third line from the image: (x+2) log 2 + (3x/(x-1)) log 3 = 2 log 3. It seems the log 27 was simplified to 3 log 3, and log 9 to 2 log 3. This means they are not taking log base 3. Let's assume base 10 or e. (x+2) log 2 + (3x / (x-1)) log 3 = 2 log 3 Divide by log 3: (x+2) (log 2 / log 3) + 3x / (x-1) = 2 (x+2) log_3 2 + 3x / (x-1) = 2 This leads back to the equation we solved which gave x = -2 or x = 1 + log_(1/2) 3. Let's check the solution's final steps in the image again. or (x+2) [ log 2 + log 3 / (x-1) ] = 0 -- this line appears incorrect. Let's assume it is meant to be (x+2) log 2 + (3x/(x-1)) log 3 = 2 log 3. Divide by log 2 instead. (x+2) + (3x/(x-1)) (log 3 / log 2) = 2 (log 3 / log 2) (x+2) + (3x/(x-1)) log_2 3 = 2 log_2 3 Let's assume the line means: (x+2) [ log 2 + (3x/(x-1)) (log 3 / log 2) ] = 2 log 3 ??? No. Let's assume the line means: (x+2) log 2 + log [ 3^(3x/(x-1)) ] = log 9. This is not an addition inside log. Let's look at the previous line: (x+2) log 2 + x/(x-1) 3 log 3 = 2 log 3. If we assume the base of log is the same everywhere (say 10). (x+2) log 2 + (3x/(x-1)) log 3 = 2 log 3. Let's rewrite this as: (x+2) log 2 = 2 log 3 - (3x/(x-1)) log 3 (x+2) log 2 = ( 2 - 3x/(x-1) ) log 3 (x+2) log 2 = ( (2x - 2 - 3x)/(x-1) ) log 3 (x+2) log 2 = ( (-x-2)/(x-1) ) log 3 (x+2) log 2 + (x+2)/(x-1) log 3 = 0 (x+2) [ log 2 + (1/(x-1)) log 3 ] = 0 (x+2) [ (x-1) log 2 + log 3 ] / (x-1) = 0 x = -2 or (x-1) log 2 + log 3 = 0 (x-1) log 2 = -log 3 log (2^(x-1)) = log (3^(-1)) 2^(x-1) = 1/3 x-1 = log_2 (1/3) = -log_2 3 x = 1 - log_2 3 Let's compare this with the result shown in the image: => x = -2 or x-1 = -log 3 / log 2 => x = -2 or x = 1 - log_2 3 This matches the derivation (x+2) [ log 2 + (1/(x-1)) log 3 ] = 0, assuming log base 10 or e. So the line (x+2) [ log 2 + log 3 / (x-1) ] = 0 is correct. It comes from factorizing (x+2) out of (x+2) log 2 + (x+2)/(x-1) log 3 = 0. The next line shows: => x = -2 or log 2 + log 3 / (x-1) = 0 log 3 / (x-1) = - log 2 log 3 = -(x-1) log 2 log 3 = (1-x) log 2 log 3 / log 2 = 1 - x log_2 3 = 1 - x x = 1 - log_2 3. This derivation looks correct now. ILLUSTRATION 1.47 Solve log_2(4 x 3^x – 6) – log_2(9^x – 6) = 1. Sol. log_2(4 x 3^x – 6) – log_2(9^x – 6) = 1 or log_2 [ (4 x 3^x – 6) / (9^x – 6) ] = 1 or (4 x 3^x – 6) / (9^x – 6) = 2^1 = 2 or 4 x 3^x – 6 = 2 (9^x – 6) or 4 x 3^x – 6 = 2 x (3^2)^x – 12 or 4 x 3^x – 6 = 2 x (3^x)^2 – 12 Let y = 3^x. or 4y – 6 = 2y^2 – 12 (putting 3^x = y) or 2y^2 – 4y – 6 = 0 or y^2 – 2y – 3 = 0 or (y – 3)(y + 1) = 0 or y = 3 or y = –1, 3^x = 3 or 3^x = –1 Since 3^x must be positive, 3^x = 3 is the only possibility. or 3^x = 3 or x = 1 For x=1, 4 * 3^1 - 6 = 12 - 6 = 6 > 0. 9^1 - 6 = 9 - 6 = 3 > 0. So the terms are defined. ILLUSTRATION 1.48 Solve 6(log_2 x – log_4 x)^2 + 7 = 0. Sol. 6(log_2 x – log_4 x)^2 + 7 = 0 or 6(log_2 x – log_2 x / log_2 4)^2 + 7 = 0 or 6(log_2 x – log_2 x / 2)^2 + 7 = 0 Let y = log_2 x. or 6(y – y/2)^2 + 7 = 0 or 6(y/2)^2 + 7 = 0 or 6(y^2/4) + 7 = 0 or 3y^2/2 + 7 = 0 or 3y^2 + 14 = 0 or y^2 = –14/3 Since y^2 cannot be negative, there is no real solution for y. Therefore, there is no real solution for x. Let's re-read the problem carefully. Is it 6(log_2 x – log_4 x)^2 + 7 = 0 or something else? Looking closely at the image, it appears to be 6(log_2 x – log_x 4)^2 + 7 = 0. Let's solve this version. Solve 6(log_2 x – log_x 4)^2 + 7 = 0. Sol. 6(log_2 x – log_x 4)^2 + 7 = 0 or 6(log_2 x – log_2 4 / log_2 x)^2 + 7 = 0 or 6(log_2 x – 2 / log_2 x)^2 + 7 = 0 Let y = log_2 x. or 6(y – 2/y)^2 + 7 = 0 or 6( (y^2 – 2)/y )^2 + 7 = 0 or 6 (y^2 – 2)^2 / y^2 + 7 = 0 or 6(y^4 – 4y^2 + 4) + 7y^2 = 0 (assuming y != 0, i.e., x != 1) or 6y^4 – 24y^2 + 24 + 7y^2 = 0 or 6y^4 – 17y^2 + 24 = 0 Let z = y^2. 6z^2 – 17z + 24 = 0 Discriminant D = (–17)^2 – 4(6)(24) = 289 – 576 = –287. Since D < 0, there are no real solutions for z = y^2. This means there are no real solutions for y, and thus no real solutions for x. Let's look at the solution steps in the image again for Illustration 1.48. Solve 6(log_2 x – log_4 x) + 7 = 0. -- The problem statement in the image header says 6(log_2 x – log_4 x)^2 + 7 = 0. But the solution starts with 6(log_2 x – log_4 x) + 7 = 0 (without the square). Let's follow the solution in the image. Sol. 6(log_2 x – log_4 x) + 7 = 0 or 6(log_2 x – (1/2)log_2 x) + 7 = 0 (where y = log_2 x is written on the right) This substitution y = log_2 x is written next to the previous line, which is confusing. Let's assume y = log_2 x. 6(y – y/2) + 7 = 0 or 6(y/2) + 7 = 0 or 3y + 7 = 0 or y = –7/3 Substitute back y = log_2 x: log_2 x = –7/3 x = 2^(-7/3) = 1 / (2^(7/3)) = 1 / (cube_root(2^7)) = 1 / (cube_root(128)). This is a valid solution since x > 0. Let's re-examine the image solution lines for 1.48. Solve 6(log_2 x – log_4 x) + 7 = 0. Sol. 6(log_2 x – log_4 x) + 7 = 0 or 6(log_2 x – 1/2 log_2 x) + 7 = 0 (where y = log_2 x) -- It seems y=log_2 x is defined here. or 6(1/2 y) + 7 = 0 -- This line seems to substitute y in the expression 6(log_2 x - 1/2 log_2 x). 6(1/2 log_2 x) + 7 = 0 3 log_2 x + 7 = 0 3 log_2 x = -7 log_2 x = -7/3 x = 2^(-7/3) Let's look at the next lines in the image: or 6((2-y^2)/y) + 7 = 0 -- This looks completely different and corresponds to the (y - 2/y) expression from the squared version. Let's assume the problem intended to be 6(log_2 x – log_x 4) + 7 = 0, not squared. 6(log_2 x – log_x 4) + 7 = 0 6(log_2 x – 2/log_2 x) + 7 = 0 Let y = log_2 x. 6(y – 2/y) + 7 = 0 6(y^2 - 2)/y + 7 = 0 6(y^2 - 2) + 7y = 0 (assuming y != 0) 6y^2 - 12 + 7y = 0 6y^2 + 7y - 12 = 0 (3y - 4)(2y + 3) = 0 3y - 4 = 0 => y = 4/3 => log_2 x = 4/3 => x = 2^(4/3) 2y + 3 = 0 => y = -3/2 => log_2 x = -3/2 => x = 2^(-3/2) Now let's look at the solution lines in the image again, starting from the line with 6((2-y^2)/y) + 7 = 0. It seems the intended problem was 6(log_2 x – log_x 4) + 7 = 0. Let y = log_2 x. log_x 4 = log_2 4 / log_2 x = 2/y. The expression is 6(y - 2/y) + 7 = 0. 6(y^2 - 2)/y + 7 = 0 6(y^2 - 2) + 7y = 0 => 6y^2 + 7y - 12 = 0. The solution in the image shows: or 6((2-y^2)/y) + 7 = 0 -- This should be 6((y^2-2)/y) + 7 = 0. There is a sign error inside the parenthesis. Let's assume the sign error is corrected: 6((y^2-2)/y) + 7 = 0 or (6(y^2-2) + 7y) / y = 0 or 6y^2 - 12 + 7y = 0 => 6y^2 + 7y - 12 = 0 The next line in the image is: or 3(y^2) - 7y + 0 = 0 -- This looks completely wrong compared to 6y^2 + 7y - 12 = 0. Let's look at the next line: or 3y^2 – 7y – 6 = 0 -- This might be a new equation. Let's try to solve this one. (3y + 2)(y - 3) = 0 => y = -2/3 or y = 3. Let's check if this fits any possible interpretation of the original problem or solution steps. Let's assume the original problem is 6(log_2 x)^2 - 7 log_2 x - 6 = 0. Let y = log_2 x. 6y^2 - 7y - 6 = 0 (3y + 2)(2y - 3) = 0 y = -2/3 or y = 3/2. log_2 x = -2/3 => x = 2^(-2/3) log_2 x = 3/2 => x = 2^(3/2) Let's look at the solution lines again, starting from the line 3y^2 – 7y – 6 = 0. or (3y + 2)(y – 3) = 0 -- This factorization is incorrect. (3y+2)(y-3) = 3y^2 - 9y + 2y - 6 = 3y^2 - 7y - 6. This IS correct. So, the equation 3y^2 - 7y - 6 = 0 leads to y = -2/3 or y = 3. The next line in the image: or y = 3 or y = –2/3 This matches the roots of 3y^2 - 7y - 6 = 0. The next line: => log_2 x = 3 or –2/3 This assumes y = log_2 x. or x = 2^3 or x = 2^(–2/3) or x = 8 or x = 2^(-2/3) Let's try to reconstruct the original problem statement based on the equation 3y^2 - 7y - 6 = 0 where y = log_2 x. This would be 3(log_2 x)^2 - 7(log_2 x) - 6 = 0. Let's check the header: 6(log_2 x – log_4 x)^2 + 7 = 0. Let y = log_2 x. log_4 x = log_2 x / log_2 4 = y/2. 6(y - y/2)^2 + 7 = 0 6(y/2)^2 + 7 = 0 6(y^2/4) + 7 = 0 3y^2/2 + 7 = 0 3y^2 + 14 = 0. This has no real solutions. There is a significant discrepancy between the header (problem statement) and the solution steps provided for Illustration 1.48. The solution steps starting from "or 3y^2 – 7y – 6 = 0" are consistent with the quadratic equation 3y^2 - 7y - 6 = 0 where y = log_2 x, leading to x=8 or x=2^(-2/3). The line "or 6(log_2 x – 1/2 log_2 x) + 7 = 0 (where y = log_2 x)" suggests the problem was 6(log_2 x – log_4 x) + 7 = 0, which yields x = 2^(-7/3). The line "or 6((2-y^2)/y) + 7 = 0" seems related to log_x 4, possibly from the squared version, but it has a sign error and doesn't lead to the final quadratic 3y^2 - 7y - 6 = 0. Given the clear sequence from 3y^2 - 7y - 6 = 0 to the final answers, it is most likely that the intended problem was one that simplifies to this quadratic in y = log_2 x. One possibility is 3(log_2 x)^2 - 7(log_2 x) - 6 = 0, but this does not match the header. Another possibility is manipulating log_2 x and log_x 4. Let y = log_2 x. log_x 4 = 2/y. Consider the expression: A (y)^2 + B (2/y) + C = 0 or A (y) + B (2/y)^2 + C = 0 or A(y)^2 + B(2/y)^2 + C = 0 etc. Or maybe a combination like (y - 2/y) multiplied by some factor, or (y - 2/y) in a quadratic. Let's assume the header is correct: 6(log_2 x – log_4 x)^2 + 7 = 0. As shown, this leads to 3y^2 + 14 = 0 (where y=log_2 x), no real solution. Let's assume the header is a typo and the problem is 6(log_2 x - log_x 4) + 7 = 0. This leads to 6y^2 + 7y - 12 = 0 (where y=log_2 x), with solutions y = 4/3, -3/2. Let's assume the header is a typo and the problem is 6(log_2 x - log_x 4)^2 + 7 = 0. This leads to 6(y - 2/y)^2 + 7 = 0, which gave 6y^4 - 17y^2 + 24 = 0 (let z=y^2, 6z^2 - 17z + 24 = 0), no real solution for y^2. Given that the solution clearly arrives at y=3 or y=-2/3 for y=log_2 x, let's assume the intended quadratic in y=log_2 x was 3y^2 - 7y - 6 = 0. The solution steps in the image jump from "6(log_2 x – 1/2 log_2 x) + 7 = 0" to "6((2-y^2)/y) + 7 = 0" which is totally inconsistent. Then it jumps to "3y^2 – 7y – 6 = 0". Conclusion for 1.48: The problem statement in the header appears inconsistent with the provided solution steps that lead to a result. However, the final steps starting from "3y^2 – 7y – 6 = 0" leading to x=8 or x=2^(-2/3) are consistent with solving 3(log_2 x)^2 - 7(log_2 x) - 6 = 0. I will extract the problem statement as given and the solution steps as given, highlighting the inconsistencies where apparent. Given the format request is for *extraction*, I will extract what is visually present. ILLUSTRATION 1.48 (As presented in the image) Solve 6(log_2 x – log_4 x)^2 + 7 = 0. Sol. 6(log_2 x – log_4 x) + 7 = 0 or 6(log_2 x – 1/2 log_2 x) + 7 = 0 (where y = log_2 x) or 6(1/2 y) + 7 = 0 or 6((2-y^2)/y) + 7 = 0 -- This line is inconsistent with the previous lines. or 3(y^2) - 7y + 0 = 0 -- This line is also inconsistent. or 3y^2 – 7y – 6 = 0 -- This is a new quadratic equation. or (3y + 2)(y – 3) = 0 or y = 3 or y = –2/3 => log_2 x = 3 or –2/3 or x = 2^3 or x = 2^(–2/3) or x = 8 or x = 2^(-2/3) ILLUSTRATION 1.49 Solve 4^(log_2 log x) = log x – (log x)^2 + 1 (base is e). Sol. log_2 log x is meaningful if log x > 0, i.e., x > e^0 = 1. Also log x should be > 0 for log_2(log x). Since 4^(log_2 log x) = 2^(2 log_2 log x) = (2^(log_2 log x))^2 = (log x)^2 (Using a^(log_a M) = M) Let y = log x (base e). The equation is y > 0. (log x)^2 = log x – (log x)^2 + 1 2(log x)^2 – log x – 1 = 0. So the given equation reduces to 2(log x)^2 – log x – 1 = 0. Let y = log x. 2y^2 – y – 1 = 0 (2y + 1)(y – 1) = 0 y = –1/2 or y = 1. log x = –1/2 or log x = 1. x = e^(-1/2) or x = e^1. We need log x > 0, which means y > 0. So y = 1 is the only valid solution for y. log x = 1 x = e. For x=e, log x = log e = 1. log_2(log x) = log_2(1) = 0. LHS = 4^(log_2 log x) = 4^0 = 1. RHS = log x – (log x)^2 + 1 = 1 – (1)^2 + 1 = 1 – 1 + 1 = 1. LHS = RHS. The solution x=e is correct. Checking the side note "base is e". This confirms log x means ln x. Checking the line "Since 4^(log_2 log x) = 2^(2 log_2 log x) = (2^(log_2 log x))^2 = (log x)^2 (a^(log_a x) = x, a > 0, a != 1)". This derivation is correct. 1.12 Trigonometry Therefore, log x = 1 or log x = –1/2. But for x > 1, log x > 0. Hence, log x = 1, i.e., x = e. This seems to be a continuation of the solution for ILLUSTRATION 1.49, reiterating the condition x > 1 (derived from log x > 0 because log_2 log x requires log x > 0, and for base e, log x > 0 means x > e^0 = 1) and selecting the valid solution. ILLUSTRATION 1.50 Solve 4log_(x/2) (sqrt(x)) + 2log_(4x) (x^2) = 3log_(2x) (x^3). Sol. Change all logs to base 2. log_(a) b = log_c b / log_c a log_(x/2) sqrt(x) = log_2 sqrt(x) / log_2 (x/2) = (1/2) log_2 x / (log_2 x - log_2 2) = (1/2) log_2 x / (log_2 x - 1) log_(4x) x^2 = log_2 x^2 / log_2 (4x) = 2 log_2 x / (log_2 4 + log_2 x) = 2 log_2 x / (2 + log_2 x) log_(2x) x^3 = log_2 x^3 / log_2 (2x) = 3 log_2 x / (log_2 2 + log_2 x) = 3 log_2 x / (1 + log_2 x) Substitute these into the equation: 4 * [ (1/2) log_2 x / (log_2 x - 1) ] + 2 * [ 2 log_2 x / (2 + log_2 x) ] = 3 * [ 3 log_2 x / (1 + log_2 x) ] Let y = log_2 x. 4 * (y/2) / (y-1) + 2 * (2y) / (2+y) = 3 * (3y) / (1+y) 2y / (y-1) + 4y / (y+2) = 9y / (y+1) The solution in the image has different denominators for the second and third terms in the first step after "Sol.". Let's check them. Sol. 4 log_2 sqrt(x) / log_2 (x/2) + 2 log_2 (x^2) / log_2 (4x) = 3 log_2 (x^3) / log_2 (2x) This is correct application of change of base formula. => [ 4 * (1/2) log_2 (x) ] / [ log_2 x – 1 ] + [ 2 log_2 (x^2) ] / [ log_2 4 + log_2 (x) ] = [ 3 log_2 (x^3) ] / [ log_2 2 + log_2 (x) ] => [ 2 log_2 (x) ] / [ log_2 x – 1 ] + [ 4 log_2 (x) ] / [ 2 + log_2 (x) ] = [ 9 log_2 (x) ] / [ 1 + log_2 (x) ] Let y = log_2 x. => 2y / (y – 1) + 4y / (2 + y) = 9y / (1 + y) y [ 2 / (y – 1) + 4 / (2 + y) – 9 / (1 + y) ] = 0 y = 0 or 2 / (y – 1) + 4 / (2 + y) = 9 / (1 + y). Case 1: y = 0. log_2 x = 0 => x = 2^0 = 1. Check domain conditions for x=1: Base x/2 = 1/2. log_(1/2) sqrt(1) = log_(1/2) 1 = 0. Defined. Base 4x = 4. log_4 (1^2) = log_4 1 = 0. Defined. Base 2x = 2. log_2 (1^3) = log_2 1 = 0. Defined. LHS = 4 * 0 + 2 * 0 = 0. RHS = 3 * 0 = 0. So x = 1 (y = 0) is a solution. Case 2: 2 / (y – 1) + 4 / (2 + y) = 9 / (1 + y), where y != 0, y != 1, y != -2, y != -1. Common denominator is (y-1)(y+2)(y+1). 2(y+2)(y+1) + 4(y-1)(y+1) = 9(y-1)(y+2) 2(y^2 + 3y + 2) + 4(y^2 - 1) = 9(y^2 + y - 2) 2y^2 + 6y + 4 + 4y^2 - 4 = 9y^2 + 9y - 18 6y^2 + 6y = 9y^2 + 9y - 18 0 = 3y^2 + 3y - 18 0 = y^2 + y - 6 0 = (y + 3)(y - 2) y = -3 or y = 2. If y = -3, log_2 x = -3 => x = 2^(-3) = 1/8. Check domain conditions for x = 1/8: Base x/2 = (1/8)/2 = 1/16 > 0 and != 1. Defined. Base 4x = 4(1/8) = 1/2 > 0 and != 1. Defined. Base 2x = 2(1/8) = 1/4 > 0 and != 1. Defined. Arguments sqrt(x), x^2, x^3 are all (1/8)^(positive power), which are positive. Defined. So x = 1/8 is a solution. If y = 2, log_2 x = 2 => x = 2^2 = 4. Check domain conditions for x = 4: Base x/2 = 4/2 = 2 > 0 and != 1. Defined. Base 4x = 4(4) = 16 > 0 and != 1. Defined. Base 2x = 2(4) = 8 > 0 and != 1. Defined. Arguments sqrt(x), x^2, x^3 are all positive. Defined. So x = 4 is a solution. The solutions are x = 1, x = 1/8, x = 4. ILLUSTRATION 1.53 Solve |x – 1|^((log_10 x)^2 – log_10 x^2) = |x – 1|^3. Sol. We have |x – 1|^((log_10 x)^2 – log_10 x^2) = |x – 1|^3. The base is |x-1|. The exponent is (log_10 x)^2 – log_10 x^2 = (log_10 x)^2 – 2 log_10 x. Let y = log_10 x. Exponent is y^2 - 2y. The equation is |x-1|^(y^2 - 2y) = |x-1|^3. For the logarithms log_10 x and log_10 x^2 to be defined, x > 0. For the bases of the powers, |x-1|, we need |x-1| > 0 and |x-1| != 1. |x-1| > 0 means x != 1. |x-1| != 1 means x-1 != 1 and x-1 != -1, so x != 2 and x != 0. So we need x > 0, x != 1, x != 2. Now solve |x-1|^(y^2 - 2y) = |x-1|^3. There are three cases for a^b = a^c: 1. Base a = 1. 1^b = 1^c is always true, provided b and c are defined. |x-1| = 1 => x-1 = 1 or x-1 = -1 => x = 2 or x = 0. We require x > 0, x != 1, x != 2. So neither x=2 nor x=0 are allowed by the log and base conditions. 2. Exponents are equal: b = c. y^2 - 2y = 3 y^2 - 2y - 3 = 0 (y - 3)(y + 1) = 0 y = 3 or y = -1. Substitute back y = log_10 x. log_10 x = 3 => x = 10^3 = 1000. Check conditions: x = 1000. x > 0 (yes), x != 1 (yes), x != 2 (yes). This is a valid solution. log_10 x = -1 => x = 10^(-1) = 0.1. Check conditions: x = 0.1. x > 0 (yes), x != 1 (yes), x != 2 (yes). This is a valid solution. 3. Base a = -1. (-1)^b = (-1)^c requires b and c to be integers and of the same parity. |x-1| = -1 is not possible for real x. 4. Base a = 0. 0^b = 0^c requires b > 0 and c > 0. |x-1| = 0 => x = 1. This is not allowed by the log conditions. So, the solutions are from case 2: x = 1000 and x = 0.1. Let's check the solution steps in the image: |x – 1|^((log_10 x)^2 – log_10 x^2) = |x – 1|^3, x > 0, x != 1 -- This is partially correct. It misses x != 2. => |x – 1| = 1 or (log_10 x)^2 – log_10 x^2 = 3 The first part |x-1|=1 gives x=2 or x=0. As discussed, neither is valid due to domain requirements (x>0, x!=1). The second part (log_10 x)^2 – log_10 x^2 = 3. => (log_10 x)^2 – 2 log_10 x = 3 Let y = log_10 x. y^2 – 2y – 3 = 0 => (log_10 x – 3)(log_10 x + 1) = 0 -- This factorization is incorrect. It should be (y-3)(y+1)=0, so (log_10 x - 3)(log_10 x + 1) = 0. This is correct. => log_10 x = 3 or log_10 x = –1 => x = 10^3 or x = 10^(-1) => x = 1000 or x = 0.11 -- Typo in the last value. It should be 0.1. Let's re-evaluate case 1 again, the condition |x-1|=1. If x=2, |x-1|=1. logs are log_10 2 and log_10 2^2 = 2 log_10 2. The exponent is (log_10 2)^2 - 2 log_10 2. This is defined. The base is 1. So 1^(defined exponent) = 1. RHS is |2-1|^3 = 1^3 = 1. So x=2 *is* a solution unless the rule for a^b = a^c when a=1 only applies when b and c are equal. However, 1^b = 1^c is true for any defined b, c. If x=0, logs are not defined. So x=0 is not a solution. The problem statement has |x-1| as the base. For a^b=a^c, we have: 1. a=1 => 1^b = 1^c (requires b, c defined) 2. a=0 => 0^b = 0^c (requires b > 0, c > 0) 3. a=-1 => (-1)^b = (-1)^c (requires b, c defined integers of same parity) 4. a > 0 and a != 1 => b = c In our case, a = |x-1|, b = (log_10 x)^2 – 2 log_10 x, c = 3. Logs require x > 0. So x-1 > -1. |x-1| can be 0 (at x=1, not allowed), between 0 and 1 (0 < |x-1| < 1), equal to 1 (|x-1|=1), or greater than 1 (|x-1| > 1). The condition log_10 x defined requires x > 0. The exponent b = (log_10 x)^2 – 2 log_10 x is defined for x > 0. The exponent c = 3 is always defined. Case 1: |x-1| = 1. x - 1 = 1 => x = 2. Base is |2-1| = 1. Logs are log_10 2. Exponent b = (log_10 2)^2 - 2 log_10 2. Exponent c = 3. Both are defined. 1^b = 1^c is 1 = 1. So x = 2 is a solution. x - 1 = -1 => x = 0. Logs are not defined for x=0. Not a solution. Case 2: |x-1| = 0. x = 1. Logs are not defined for x=1. Not a solution. Case 3: |x-1| > 0 and |x-1| != 1. This means x > 0, x != 1, and x != 2. In this case, b = c. (log_10 x)^2 – 2 log_10 x = 3 (log_10 x)^2 – 2 log_10 x – 3 = 0 Let y = log_10 x. y^2 - 2y - 3 = 0 => (y-3)(y+1) = 0. y = 3 or y = -1. log_10 x = 3 => x = 1000. Checks conditions: x=1000 > 0, != 1, != 2. Valid. log_10 x = -1 => x = 0.1. Checks conditions: x=0.1 > 0, != 1, != 2. Valid. Case 4: |x-1| = -1 (Not possible for real x). So the solutions are x = 2, x = 1000, x = 0.1. The solution in the image misses x=2. It also writes x = 0.11 instead of 0.1 at the end. The derivation step "|x – 1| = 1 or (log_10 x)^2 – log_10 x^2 = 3" is incomplete for the a^b=a^c case. It covers a=1 or b=c, but often you also need to consider a=-1 (requires integer exponents of same parity) and a=0 (requires positive exponents). In this specific case, a=|x-1| >= 0, so a=-1 is not possible. a=0 implies x=1 which is excluded by log definition. Extracting exactly what is in the image: ILLUSTRATION 1.53 Solve |x – 1|^((log_10 x)^2 – log_10 x^2) = |x – 1|^3. Sol. We have |x – 1|^((log_10 x)^2 – log_10 x^2) = |x – 1|^3, x > 0, x != 1 => |x – 1| = 1 or (log_10 x)^2 – log_10 x^2 = 3 => x = 2 or 0 or (log_10 x)^2 – 2 log_10 x = 3 => x = 2 or 0 or (log_10 x – 3)(log_10 x + 1) = 0 => x = 2 or 0 or log_10 x = 3, log_10 x = –1 => x = 2, x = 1000 or x = 0.11 There's a typo in the second line: x = 2 or 0 should be x=2 or x=0, derived from |x-1|=1. However, x=0 makes logs undefined. So only x=2 is potentially from this branch. There's a typo in the final line: x = 0.11 should be x = 0.1. The exclusion condition x != 1 is given, but x > 0 is only implicitly handled by log. The exclusion x != 2 is missed in the initial domain description but derived from |x-1|!=1. However, x=2 is presented as a potential solution from |x-1|=1. This indicates the solution is not handling the base conditions systematically. The correct set of solutions is x=0.1, x=2, x=1000. Final check of all sections. Looks complete based on visible text.