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

The answer to 4^{x+3}=7^{x-1} is x=(ln(448))/(ln(7)-2ln(2))

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

The solution is x=(ln(448))/(ln(7)-2ln(2))

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

x = \frac{ln ( 448 )}{ln ( 7 ) - 2 ln ( 2 )}

Solution steps

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

Apply exponent rules

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

Solve

(x + 3) ln (4) = (x - 1) ln (7) : x = \frac{ln ( 448 )}{ln ( 7 ) - 2 ln ( 2 )}

x = \frac{ln ( 448 )}{ln ( 7 ) - 2 ln ( 2 )}

7 e^{5 t} = 100

x - e^{x} = 0

4^{2} \cdot 4^{n} = 4^{7}

4^{x - 1} - 2^{x + 3} + 28 = 0

6^{x - 10} = 4

What is the answer to 4^{x+3}=7^{x-1} ?
The answer to 4^{x+3}=7^{x-1} is x=(ln(448))/(ln(7)-2ln(2))
Solve for x: 4^{x+3}=7^{x-1}
The solution is x=(ln(448))/(ln(7)-2ln(2))