Question

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

Answer 1

`x = 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) = (3^(log_3 4))^x`

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

`3^(x + 7) = 3^(x*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*log_3 4)`

The logarithms will cancel with the exponentials, leaving us with

`x + 7 = x*log_3 4`

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

`7 = xlog_3 4 - x`

Factor `x`:

`7 = x(log_3 4 - 1)`

And then divide:

`7/(log_3 4 - 1) = x`

And there is our answer.