What is the answer to 4^{2x-1}=7^{x-3} ?

The answer to 4^{2x-1}=7^{x-3} is x=(-3ln(7)+2ln(2))/(4ln(2)-ln(7))

Solve for x: 4^{2x-1}=7^{x-3}

The solution is x=(-3ln(7)+2ln(2))/(4ln(2)-ln(7))

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4^{2 x - 1} = 7^{x - 3}

x = \frac{- 3 ln ( 7 ) + 2 ln ( 2 )}{4 ln ( 2 ) - ln ( 7 )}

Decimal Notation

x = - 5.38472 \dots

Solution steps

4^{2 x - 1} = 7^{x - 3}

Apply exponent rules

(2 x - 1) ln (4) = (x - 3) ln (7)

Solve

(2 x - 1) ln (4) = (x - 3) ln (7) : x = \frac{- 3 ln ( 7 ) + 2 ln ( 2 )}{4 ln ( 2 ) - ln ( 7 )}

x = \frac{- 3 ln ( 7 ) + 2 ln ( 2 )}{4 ln ( 2 ) - ln ( 7 )}

4^{2 x} - 2 \cdot 4^{x + 1} + 16 = 0

3^{4 x + 8} = 1

e^{2 x} - e^{x} = 2

9^{x + 6} = 3^{3 x}

\frac{1}{2} = e^{- x}

What is the answer to 4^{2x-1}=7^{x-3} ?
The answer to 4^{2x-1}=7^{x-3} is x=(-3ln(7)+2ln(2))/(4ln(2)-ln(7))
Solve for x: 4^{2x-1}=7^{x-3}
The solution is x=(-3ln(7)+2ln(2))/(4ln(2)-ln(7))