What is the answer to 2^{x-1}+20=3(2^x) ?

The answer to 2^{x-1}+20=3(2^x) is x=3

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2^{x-1}+20=3(2^x)

2^{x - 1} + 20 = 3 (2^{x})

x = 3

Solution steps

2^{x - 1} + 20 = 3 (2^{x})

Subtract

3 \cdot 2^{x}

from both sides

2^{x - 1} + 20 - 3 \cdot 2^{x} = 3 \cdot 2^{x} - 3 \cdot 2^{x}

Simplify

2^{x - 1} + 20 - 3 \cdot 2^{x} = 0

Factor

2^{x - 1} + 20 - 3 \cdot 2^{x} : \frac{5}{2} (- 2^{x} + 8)

\frac{5}{2} (- 2^{x} + 8) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

- 2^{x} + 8 = 0

Solve

- 2^{x} + 8 = 0 : x = 3

x = 3

\frac{13}{2 ^{x}} = 3^{x - 5}

2^{3 x + 5} = 32 (8)^{x}

x, e^{x} = 1

2^{4 x} = 127

x, e^{x} = 7

What is the answer to 2^{x-1}+20=3(2^x) ?
The answer to 2^{x-1}+20=3(2^x) is x=3
Solve for x: 2^{x-1}+20=3(2^x)
The solution is x=3