# How do you solve 3^(x+7)=4^x?

Sep 11, 2015

$x = \frac{7}{{\log}_{3} 4 - 1}$

or, alternatively,

x = -7ln3/(ln(3/4)

#### Explanation:

When I see these types of problems, it's helpful for me to rewrite the equation such that the bases of the two exponentials are the same. What do I mean by this?

Well, the logarithm and the exponential are inverse functions, right? So, it's logical that 4 is actually the same thing as ${3}^{{\log}_{3} 4}$.

Using this logic we can write the original equation as

${3}^{x + 7} = {\left({3}^{{\log}_{3} 4}\right)}^{x}$

And now, using some laws of exponentials, we can simplify the above equation as

${3}^{x + 7} = {3}^{x \cdot {\log}_{3} 4}$

Now, how does this help us? Well, we can now take the base-3 logarithm of both sides of the equation:

${\log}_{3} {3}^{x + 7} = {\log}_{3} {3}^{x \cdot {\log}_{3} 4}$

The logarithms will cancel with the exponentials, leaving us with

$x + 7 = x \cdot {\log}_{3} 4$

From here, we just need some simple algebra to solve for $x$:

$7 = x {\log}_{3} 4 - x$

Factor $x$:

$7 = x \left({\log}_{3} 4 - 1\right)$

And then divide:

$\frac{7}{{\log}_{3} 4 - 1} = x$