# logarithmic question

• November 18th 2010, 08:25 PM
math321
logarithmic question
Code:

```need some help urgently Solve the equation. (16/9)^{x-1} = (3/4)^{x-1}```
• November 18th 2010, 09:34 PM
Jhevon
Quote:

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?
• November 18th 2010, 10:06 PM
math321
So am i finding the log 16/9 and log 3/4 and multiply it by x-1 and solve for x
• November 18th 2010, 10:16 PM
Jhevon
Quote:

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.
• November 18th 2010, 11:02 PM
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
• November 18th 2010, 11:13 PM
Jhevon
Quote:

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! (Yes)