Solve:

4 + log(base 3)(2 - x) = 9

I got:

log(base 3)(2 - x) = 5

2 - x = 3^5

2 - x = 243

-x = 241

x = -241

Don't think this is right can someone help me out? Thanks.

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- Nov 12th 2013, 06:50 PMStephenSkinnerLogarithms Question HELP
Solve:

4 + log(base 3)(2 - x) = 9

I got:

log(base 3)(2 - x) = 5

2 - x = 3^5

2 - x = 243

-x = 241

x = -241

Don't think this is right can someone help me out? Thanks. - Nov 12th 2013, 07:16 PMSlipEternalRe: Logarithms Question HELP
Your question was to solve $\displaystyle 4+\log_3(2-x) = 5$ and you solved $\displaystyle \log_3(2-x) = 5$. Those are not the same equations. To solve the first equation, subtract 4 from each side:

$\displaystyle \begin{align*}& \log_3(2-x) = 1 \\ \Rightarrow & 2-x = 3^1 = 3 \\ \Rightarrow & x = -1\end{align*}$. - Nov 12th 2013, 07:17 PMProve ItRe: Logarithms Question HELP
Surely 5 - 4 is NOT 5.

- Nov 12th 2013, 07:31 PMStephenSkinnerRe: Logarithms Question HELP
Whoops meant to put: 4 + log(base 3)(2-x) = 9

SORRY. - Nov 12th 2013, 07:43 PMSlipEternalRe: Logarithms Question HELP
- Nov 12th 2013, 07:48 PMStephenSkinnerRe: Logarithms Question HELP