1. ## logarithmic question

Code:
need some help urgently

Solve the equation.
(16/9)^{x-1} = (3/4)^{x-1}

2. Originally Posted by math321
need some help urgently

Solve the equation.
(16/9)^{x-1} = (3/4)^{x-1}
log both sides so that you can bring down the powers.

$\displaystyle \log (16/9)^{x - 1} = \log (3/4)^{x - 1}$

$\displaystyle \Rightarrow (x - 1) \log (16/9) = (x - 1) \log (3/4)$

now continue (note that $\displaystyle \log (16/9)$ and $\displaystyle \log (3/4)$ are constants). where can you go from there?

3. So am i finding the log 16/9 and log 3/4 and multiply it by x-1 and solve for x

4. Originally Posted by math321
So am i finding the log 16/9 and log 3/4 and multiply it by x-1 and solve for x
yes, but just leave them as log (16/9) and log (3/4), no need to evaluate them. so you're on the right track. multiplying out and getting all x's on one side would be the way to proceed.

5. can u verify this anser for me please

(16/9)^(x-1) = (9/16)^-(x-1) = (9/16)^(1-x) = ((3/4)^2)^(1-x) = (3/4)^(2(1-x))

(3/4)^(2(1-x)) = (3/4)^(x-1)

2(1-x) = (x-1)
2-2x = x-1
3 = 3x
x = 1

6. Originally Posted by math321
can u verify this anser for me please

(16/9)^(x-1) = (9/16)^-(x-1) = (9/16)^(1-x) = ((3/4)^2)^(1-x) = (3/4)^(2(1-x))

(3/4)^(2(1-x)) = (3/4)^(x-1)

2(1-x) = (x-1)
2-2x = x-1
3 = 3x
x = 1
yes, that works!