What is the answer to 2^x+3*4^2=128 ?

The answer to 2^x+3*4^2=128 is x=(ln(80))/(ln(2))

Solve for x: 2^x+3*4^2=128

The solution is x=(ln(80))/(ln(2))

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2^{x} + 3 \cdot 4^{2} = 128

x = \frac{ln ( 80 )}{ln ( 2 )}

Decimal Notation

x = 6.32192 \dots

Solution steps

2^{x} + 3 \cdot 4^{2} = 128

Move

3 \cdot 4^{2}

to the right side

2^{x} = 128 - 3 \cdot 4^{2}

Simplify

2^{x} = 80

Apply exponent rules

x ln (2) = ln (80)

Solve

x ln (2) = ln (80) : x = \frac{ln ( 80 )}{ln ( 2 )}

x = \frac{ln ( 80 )}{ln ( 2 )}

(8^{x}) = e^{x + 7}

e^{l n (2 x)} = 12

2^{x} = 2^{1 - x}

3^{∣ 3 x - 4 ∣} = 9^{2 x - 2}

2^{2 x + 1} + 5 (2^{x}) - 3 = 0

What is the answer to 2^x+3*4^2=128 ?
The answer to 2^x+3*4^2=128 is x=(ln(80))/(ln(2))
Solve for x: 2^x+3*4^2=128
The solution is x=(ln(80))/(ln(2))