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Ber [7]
3 years ago
9

Solve f (x) = 32^x+6•1/2 = 8^x-1

Mathematics
1 answer:
emmainna [20.7K]3 years ago
6 0

Option B: The value of x is -16

Explanation:

Given that the equation f(x)=32^{x+6} \cdot \frac{1}{2}=8^{x-1}

We need to determine the value of x.

Let us substitute f(x) = 0, then we have,

32^{x+6} \cdot \frac{1}{2}=8^{x-1}

Now, we shall determine the value of x.

The term 8^{x-1} can be written as \left(2^{3}\right)^{x-1}

Hence, we have,

32^{x+6} \cdot \frac{1}{2}=\left(2^{3}\right)^{x-1}

Also, the term 32^{x+6} can be written as \left(2^{5}\right)^{x+6}

Thus, we have,

\left(2^{5}\right)^{x+6} \frac{1}{2}=\left(2^{3}\right)^{x-1}

Applying the exponent rule, \left(a^{b}\right)^{c}=a^{b c}, we have,

   2^{5(x+6)} \cdot \frac{1}{2}=2^{3(x-1)}

2^{5(x+6)} \cdot 2^{-1}=2^{3(x-1)}

If a^{f(x)}=a^{g(x)} then f(x)=g(x)

5(x+6)-1=3(x-1)

Simplifying, we get,

5x+30-1=3x-3

     5x+29=3x-3

     2x+29=-3

             2x=-32

               x=-16

Therefore, the value of x is -16

Hence, Option B is the correct answer.

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