What is the answer to 2^x*4^{1-x}=5^{2-x} ?

The answer to 2^x*4^{1-x}=5^{2-x} is x=(2ln(5)-2ln(2))/(-ln(2)+ln(5))

Solve for x: 2^x*4^{1-x}=5^{2-x}

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

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2^{x} \cdot 4^{1 - x} = 5^{2 - x}

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

Decimal Notation

x = 2

Solution steps

2^{x} \cdot 4^{1 - x} = 5^{2 - x}

Apply exponent rules

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

Solve

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

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

e^{3 x - 6} = 14

300 \cdot 5^{0.1 x} = 1000

b, a^{b} = c

ln (e^{3 x}) + 5 = 14

(3)^{2 x + 1} - 93 = 0

What is the answer to 2^x*4^{1-x}=5^{2-x} ?
The answer to 2^x*4^{1-x}=5^{2-x} is x=(2ln(5)-2ln(2))/(-ln(2)+ln(5))
Solve for x: 2^x*4^{1-x}=5^{2-x}
The solution is x=(2ln(5)-2ln(2))/(-ln(2)+ln(5))