What is the answer to 2^{2-x}=5^{x+1} ?

The answer to 2^{2-x}=5^{x+1} is x=-(ln(5)-2ln(2))/(ln(10))

Solve for x: 2^{2-x}=5^{x+1}

The solution is x=-(ln(5)-2ln(2))/(ln(10))

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2^{2 - x} = 5^{x + 1}

x = - \frac{ln ( 5 ) - 2 ln ( 2 )}{ln ( 10 )}

Decimal Notation

x = - 0.09691 \dots

Solution steps

2^{2 - x} = 5^{x + 1}

Apply exponent rules

(2 - x) ln (2) = (x + 1) ln (5)

Solve

(2 - x) ln (2) = (x + 1) ln (5) : x = - \frac{ln ( 5 ) - 2 ln ( 2 )}{ln ( 10 )}

x = - \frac{ln ( 5 ) - 2 ln ( 2 )}{ln ( 10 )}

2^{n} = 100

8 \cdot 2^{x + 1} = 2^{2 x}

\frac{1}{6 ^{x - 1}} = 216

5^{2 x} = 8

e^{2 x} - 13 e^{x} + 42 = 0

What is the answer to 2^{2-x}=5^{x+1} ?
The answer to 2^{2-x}=5^{x+1} is x=-(ln(5)-2ln(2))/(ln(10))
Solve for x: 2^{2-x}=5^{x+1}
The solution is x=-(ln(5)-2ln(2))/(ln(10))