What is the answer to 2(3^{(4x-5)})=7^{2x} ?

The answer to 2(3^{(4x-5)})=7^{2x} is x=(-ln(2)+5ln(3))/(2(-ln(7)+2ln(3)))

Solve for x: 2(3^{(4x-5)})=7^{2x}

The solution is x=(-ln(2)+5ln(3))/(2(-ln(7)+2ln(3)))

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2 (3^{(4 x - 5)}) = 7^{2 x}

x = \frac{- ln ( 2 ) + 5 ln ( 3 )}{2 ( - ln ( 7 ) + 2 ln ( 3 ) )}

Decimal Notation

x = 9.54961 \dots

Solution steps

2 (3^{(4 x - 5)}) = 7^{2 x}

Apply exponent rules

ln (2) + (4 x - 5) ln (3) = 2 x ln (7)

Solve

ln (2) + (4 x - 5) ln (3) = 2 x ln (7) : x = \frac{- ln ( 2 ) + 5 ln ( 3 )}{2 ( - ln ( 7 ) + 2 ln ( 3 ) )}

x = \frac{- ln ( 2 ) + 5 ln ( 3 )}{2 ( - ln ( 7 ) + 2 ln ( 3 ) )}

(5^{x})^{3} \cdot 25 = 1

4^{2 x - 5} = 2^{3 x + 7}

8 = 4 e^{10 x}

4^{(x + 4)} - 4^{(x + 3)} = 96

e^{x} - 6 = 0

What is the answer to 2(3^{(4x-5)})=7^{2x} ?
The answer to 2(3^{(4x-5)})=7^{2x} is x=(-ln(2)+5ln(3))/(2(-ln(7)+2ln(3)))
Solve for x: 2(3^{(4x-5)})=7^{2x}
The solution is x=(-ln(2)+5ln(3))/(2(-ln(7)+2ln(3)))