solving an equation with different base

I am trying to solve a equation $(2^x)(3^{x-2}) = 4$ (the solution is $x = 2$)

My approach is to use the natural logarithm $\ln$ (that's suggested when googleing the problem)

So this is how it goes for me:

$$\ln{2^x}\cdot\ln{3^{x-2}} = \ln{2^2}$$

but I am unsure if using $\ln$ is correct or where to go from here.

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Thank you for all of the replies i really appreciate the help! I can see two types of approach one using logarithms and the other is just using exponents... which is more correct ? – 54N1 Nov 4 '12 at 17:38
All of them works, it depends on what you can use and what you feel more confortable with – Jean-Sébastien Nov 4 '12 at 18:27

3 Answers

The correct way to use the logarithm is to take the logarithm of both sides:

$$\ln{(2^{x}\cdot3^{x-2})}=\ln{4}$$

Using the logarithm law: $\ln{a\cdot b}=\ln{a}+\ln{b}$, we now have:

$$\ln{2^{x}}+\ln{3^{x-2}}=\ln{4}$$

We now use another law of logarithms: $\ln{a^{b}}=b\ln{a}$:

$$x\ln{2}+(x-2)\ln{3}=2\ln{2}$$

Re-arranging we get:

$$x(\ln{2}+\ln{3})=2\ln{2}+2\ln{3}$$

We can solve for $x$ to get:

$$x=\frac{2(\ln{2}+\ln{3})}{\ln{2}+\ln{3}}=2$$

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Note that $2^x 3^{x-2} = \dfrac{6^x}9$. This gives us that $6^x = 36 \implies x = 2$.

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$(2^x)(3^{x-2}) = 4$

$2^x 3^x3^{-2} = 4$

$6^x/3^2 = 4$

$6^x =36$

$6^x =6^2$

$x=2$

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