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

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

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

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

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

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

Decimal Notation

x = - 0.87823 \dots

Solution steps

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

Apply exponent rules

x ln (4) = (2 x + 1) ln (5)

Solve

x ln (4) = (2 x + 1) ln (5) : x = \frac{ln ( 5 )}{2 ( ln ( 2 ) - ln ( 5 ) )}

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

x^{x^{3}} = 2^{2}

x, e^{x} + e^{- x} = 3

k, n = 2^{k}

3^{x} = 19

9^{b^{3} - 1} = 81

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