Solving for the Exponent

We can see that simply taking the natural logarithm of both sides does not yield a solution, so to solve this, we must first see that:

$2^{x - 4}$ = $\dfrac{2^x}{2^4}$ = $\dfrac{2^x}{16}$

Then:

$\dfrac{2^x}{16} - 2^x = -15$

We can then multiply both sides by $16$ :

$2^x - 16(2^x) = -240$

We can then see that we have like terms, with different coefficients, so we can now subtract the two expressions:

$-15(2^x) = -240$

Then:

$2^x = 16$

We can then take the logarithm of both sides, with base $2$ , or we can take the natural logarithm of both sides with the base change and see that:

$x = 4$

$2^{x-4}= \frac {2^x}{2^4}$ so we can rewrite the equation as $\frac{2^x}{2^4}- 2^x = -15$

If we then take $2^x = y$ we can simply solve for $y$ .

$\frac {y}{2^4} - y = -15$

$\frac {y}{16} - y = -15$

Factorising $y$ we get $( \frac {1}{16} - 1 ) y = -15$

$\frac {-15}{16} y = -15$

Divide by $-15$ on both sides

$\frac {1}{16} y = 1$

Multiply by 16 on both sides

$y = 16$

$y = 2^x$

$\Rightarrow 2^x = 16$

$log_2 2^x = log_2 16$

$x = 4$

$2^{x-4} - 2^x = -15$

$2^x*2^{-4} - 2^x = -12$

$2^x(2^{-4} - 1) = -15$

$2^x = 16$

$2^x = 2^4$

$x = 4$