1
$\begingroup$

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.

  • 0
    All of them works, it depends on what you can use and what you feel more confortable with2012-11-04

3 Answers 3

3

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

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

$6^x/3^2 = 4$

$6^x =36$

$6^x =6^2$

$x=2$

4

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

3

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:

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