1. ## Logarithms 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.

2. ## Re: 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*}.

3. ## Re: Logarithms Question HELP

Surely 5 - 4 is NOT 5.

4. ## Re: Logarithms Question HELP

Whoops meant to put: 4 + log(base 3)(2-x) = 9

SORRY.

5. ## Re: Logarithms Question HELP

Originally Posted by StephenSkinner
Whoops meant to put: 4 + log(base 3)(2-x) = 9

SORRY.

$\displaystyle 4+\log_3(2-(-241)) = 4+\log_3(243) = 4+\log_3(3^5) = 4+5\log_3(3) = 4+5(1) = 9$

It looks correct to me.

6. ## Re: Logarithms Question HELP

Originally Posted by SlipEternal

$\displaystyle 4+\log_3(2-(-241)) = 4+\log_3(243) = 4+\log_3(3^5) = 4+5\log_3(3) = 4+5(1) = 9$

It looks correct to me.
Cool I appreciate it!