Question

revisiting exponents & their functions practice
complete this assessment to review what you’ve learned. it will not

solve for ( x ) in the equation ( 8^{-x - 6} = left( \frac{1}{16}
ight)^{x - 5} ). (1 point)

( x = square )

Explanation:

Step1: Express bases as powers of 2

We know that \(8 = 2^3\) and \(\frac{1}{16}=2^{-4}\) (since \(\frac{1}{16}=\frac{1}{2^4} = 2^{-4}\)). Substitute these into the equation:
\((2^3)^{-x - 6}=(2^{-4})^{x - 5}\)

Step2: Apply exponent power rule \((a^m)^n=a^{mn}\)

For the left - hand side, \((2^3)^{-x - 6}=2^{3\times(-x - 6)}=2^{-3x-18}\)
For the right - hand side, \((2^{-4})^{x - 5}=2^{-4\times(x - 5)}=2^{-4x + 20}\)
Now our equation becomes \(2^{-3x-18}=2^{-4x + 20}\)

Step3: Set exponents equal (since bases are equal and the function \(y = 2^x\) is one - to - one)

If \(a^m=a^n\) and \(a>0,a
eq1\), then \(m = n\). So we have:
\(-3x-18=-4x + 20\)

Step4: Solve for x

Add \(4x\) to both sides of the equation:
\(-3x + 4x-18=-4x+4x + 20\)
\(x-18 = 20\)
Add 18 to both sides:
\(x-18 + 18=20 + 18\)
\(x=38\)

Answer:

\(x = 38\)