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

The answer to solvefor x,2^{x+1}=5^{1-2x} is x=(ln(5)-ln(2))/(ln(50))

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

solve for

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

x = \frac{ln ( 5 ) - ln ( 2 )}{ln ( 50 )}

Decimal Notation

x = 0.23422 \dots

Solution steps

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

Apply exponent rules

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

Solve

(x + 1) ln (2) = (1 - 2 x) ln (5) : x = \frac{ln ( 5 ) - ln ( 2 )}{ln ( 50 )}

x = \frac{ln ( 5 ) - ln ( 2 )}{ln ( 50 )}

5^{6} \cdot 5^{m} = 5^{2 m} \cdot 5

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

(5^{x})^{3} = 25 \cdot 12 5^{x}

5^{b} \cdot 5^{b} = 5^{2 b}

0.5 \cdot e^{4 x} = 13

What is the answer to solvefor x,2^{x+1}=5^{1-2x} ?
The answer to solvefor x,2^{x+1}=5^{1-2x} is x=(ln(5)-ln(2))/(ln(50))