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

The answer to 7(2^x)=2(3^{4x-4}) is x=(ln(567)-ln(2))/(4ln(3)-ln(2))

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

The solution is x=(ln(567)-ln(2))/(4ln(3)-ln(2))

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

x = \frac{ln ( 567 ) - ln ( 2 )}{4 ln ( 3 ) - ln ( 2 )}

Decimal Notation

x = 1.52573 \dots

Solution steps

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

Apply exponent rules

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

Solve

ln (7) + x ln (2) = ln (2) + (4 x - 4) ln (3) : x = \frac{ln ( 567 ) - ln ( 2 )}{4 ln ( 3 ) - ln ( 2 )}

x = \frac{ln ( 567 ) - ln ( 2 )}{4 ln ( 3 ) - ln ( 2 )}

4 (3^{x}) = 3 (6^{3 x - 3})

3^{3 x} = 7^{2 x + 1}

4^{x} = 8

2^{1 - x^{2}} = \frac{1}{8}

e^{2 x} - e^{x} - 6 = 0

What is the answer to 7(2^x)=2(3^{4x-4}) ?
The answer to 7(2^x)=2(3^{4x-4}) is x=(ln(567)-ln(2))/(4ln(3)-ln(2))
Solve for x: 7(2^x)=2(3^{4x-4})
The solution is x=(ln(567)-ln(2))/(4ln(3)-ln(2))