Code:`need some help urgently`

Solve the equation.

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

Printable View

- Nov 18th 2010, 08:25 PMmath321logarithmic questionCode:
`need some help urgently`

Solve the equation.

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

- Nov 18th 2010, 09:34 PMJhevon
log both sides so that you can bring down the powers.

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

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

now continue (note that $\displaystyle \displaystyle \log (16/9)$ and $\displaystyle \displaystyle \log (3/4)$ are constants). where can you go from there? - Nov 18th 2010, 10:06 PMmath321
So am i finding the log 16/9 and log 3/4 and multiply it by x-1 and solve for x

- Nov 18th 2010, 10:16 PMJhevon
- Nov 18th 2010, 11:02 PMmath321
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 - Nov 18th 2010, 11:13 PMJhevon