# Equation with Logarithm

I want to solve the following equation:

$$3^x3^{x-1} = 243.$$

My approach is the following:

$3^{2x-1} = 243$ then:

$(2x-1)\cdot\log3 = \log 243$ and then:

$x = (\frac{\log243}{\log3}+1)/2$

Is this correct?

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Looks good to me! – icurays1 Apr 25 '13 at 7:41
Yes that is correct. what you have is equal to $3$. Also note that $3^5 =243$ – Gorg Apr 25 '13 at 7:42

This is correct. Alternatively you could use the base $3$ logarithm, in which case you obtain $$x = \frac{1}{2}(\log_3 243+1).$$ Now note that $243 = 3^5$ so this reduces to $x = \frac{1}{2}(5 + 1) = 3$.
If instead you use your expression for $x$, then you have $$x = \frac{1}{2}\left(\frac{\log 243}{\log 3} + 1\right) = \frac{1}{2}\left(\frac{\log 3^5}{\log 3} + 1\right) = \frac{1}{2}\left(\frac{5\log 3}{\log 3} + 1\right) = \frac{1}{2}(5 + 1) = 3.$$
don't need to take log cause RHS also in same base .$$3^{2x-1}=243$$ $$3^{2x-1}=3^5\implies 2x-1=5\implies x=3$$.By the way you were going on right track