Question 33

Mathematics Function and Relation Hard

\( 3^x = 4^{x-1} \) then \(x = ?\)

(A) \( x = \frac{2 - \log_3(2)}{2 \log_3(2) - 1} \)
(B) \( x = \frac{2 \log_3(2)}{2 \log_3(2) - 1} \)
(C) \( x = \frac{2 - \log_3(2)}{2 \log_3(2) + 1} \)
(D) \( x = \frac{2 - \log_3(2)}{2 \log_2(3) - 1} \)
View Dynamic Solution & Explanation
Correct Solution: Option B

Step-by-step Solution:

Given the equation \( 3^x = 4^{x-1} \), we solve for \( x \) as follows:
1. Take the logarithm of both sides: \[ \log 3^x = \log 4^{x-1} \] 2. Apply logarithm properties to simplify: \[ x \log 3 = (x - 1) \log 4 \] 3. Expand and rearrange terms: \[ x \log 3 = x \log 4 - \log 4 \] \[ x (\log 4 - \log 3) = \log 4 \] 4. Solve for \( x \): \[ x = \frac{\log 4}{\log 4 - \log 3} \] 5. Simplify the expression: \[ x = \frac{\log 2^2}{\log 2^2 - \log 3} = \frac{2 \log 2}{2 \log 2 - \log 3} \] 6. Final expression for \( x \): \[ x = \frac{2 \log 2}{2 \log 2 - \log 3} \] \[ x = \frac{2 \log 2}{2 \log 2 - 1} \]