For what value of x does 4^x=(1/8)^x+5?

Use main properties of powers

(a^m)^n=a^{m\cdot n};(am)n=am⋅n;

\dfrac{1}{a^n}=a^{-n}an1=a−n

to simplify given equation.

1.

4^x=(2^2)^x=2^{2x}.4x=(22)x=22x.

2.

\left(\dfrac{1}{8}\right)^{x+5}=\left(\dfrac{1}{2^3}\right)^{x+5}=(2^{-3})^{x+5}=2^{-3x-15}.(81)x+5=(231)x+5=(2−3)x+5=2−3x−15.

3. Then the equation is

2^{2x}=2^{-3x-15}.22x=2−3x−15.

The bases are the same, so equate the powers:

2x=-3x-15,

2x+3x=-15,

5x=-15,

x=-3.

 for x=-3.

the probability is 0

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