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 07:41
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3Yes that is correct. what you have is equal to $3$. Also note that $3^5 =243$ – Gorg Apr 25 '13 at 07:42
2 Answers
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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 $$x = \frac{1}{2}\left(\frac{\log 243}{\log 3} + 1\right) = \frac{1}{2}\left(\frac{\log 3^5}{\log 3} + 1\right) = \frac{1}{2}\left(\frac{5\log 3}{\log 3} + 1\right) = \frac{1}{2}(5 + 1) = 3.$$
Michael Albanese
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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
iostream007
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