Question 40284
Hello!
You must rewrite your equation so that the base number is equal on both sides of the equation. In your equation, on the left side, the base is 16, while on the thr right side, it's 4. However, notice that you could rewrite 16 as 4^2. So you would get:

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

However, using the rule that says that {{{ (A^B)^C = A^(B*C)}}}, we get:

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

Now, since the base is equal in both case, we simply remove it to get the equation:

{{{2x = x^2 - 15}}}
{{{ 0 = x^2 -2x -15}}}

And now we simply solve a quadratic equation:
*[invoke solve_quadratic_equation 1, -2, -15]

So we get that the two solutions are X = 5 and X = -3


I hope this helps!
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