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

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

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

The solution is x=(2ln(2))/(ln(5)-4ln(2))

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

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

Decimal Notation

x = - 1.19184 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

b, e^{3 b} = 2 + b

2^{x} + \frac{4}{2 ^{x}} = 5

500 \cdot 2^{d} = 50 \cdot 3^{d}

3 e^{- 2 x} = 75

8 1^{2^{x + 7}} = 2^{8^{x - 13}}

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