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

The answer to 4^{x+4}=7^x is x=(8ln(2))/(ln(7)-2ln(2))

Solve for x: 4^{x+4}=7^x

The solution is x=(8ln(2))/(ln(7)-2ln(2))

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4^{x + 4} = 7^{x}

x = \frac{8 ln ( 2 )}{ln ( 7 ) - 2 ln ( 2 )}

Decimal Notation

x = 9.90890 \dots

Solution steps

4^{x + 4} = 7^{x}

Apply exponent rules

(x + 4) ln (4) = x ln (7)

Solve

(x + 4) ln (4) = x ln (7) : x = \frac{8 ln ( 2 )}{ln ( 7 ) - 2 ln ( 2 )}

x = \frac{8 ln ( 2 )}{ln ( 7 ) - 2 ln ( 2 )}

2^{4 x + 1} + 3 = 7 \cdot 4^{x}

x, 2 - e^{x} = 0

3 \cdot 4^{x} = 5 \cdot 7^{2 x - 2}

7^{x} = 100

(\frac{4}{5})^{x} = 4

What is the answer to 4^{x+4}=7^x ?
The answer to 4^{x+4}=7^x is x=(8ln(2))/(ln(7)-2ln(2))
Solve for x: 4^{x+4}=7^x
The solution is x=(8ln(2))/(ln(7)-2ln(2))