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

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

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

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

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

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

Decimal Notation

x = - 2.21317 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

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

3^{2 x + 1} = 3^{π x + 4}

e^{x} + x - 1 = 0

5^{6 x} - 6 (12 5^{x}) + 5 = 0

4 \cdot 2^{x} = 2^{6 x + 3}

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