What is the answer to 5^{3x}=33(7^x) ?

The answer to 5^{3x}=33(7^x) is x=(ln(33))/(3ln(5)-ln(7))

Solve for x: 5^{3x}=33(7^x)

The solution is x=(ln(33))/(3ln(5)-ln(7))

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5^{3 x} = 33 (7^{x})

x = \frac{ln ( 33 )}{3 ln ( 5 ) - ln ( 7 )}

Decimal Notation

x = 1.21305 \dots

Solution steps

5^{3 x} = 33 (7^{x})

Apply exponent rules

3 x ln (5) = ln (33) + x ln (7)

Solve

3 x ln (5) = ln (33) + x ln (7) : x = \frac{ln ( 33 )}{3 ln ( 5 ) - ln ( 7 )}

x = \frac{ln ( 33 )}{3 ln ( 5 ) - ln ( 7 )}

3^{x} + x = 30

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

e^{2 x} = e^{x} + 6

\frac{100}{1 + 15 e ^{- 0.4 x}} = 20

3^{x - 4} = 8^{2 x + 2}

What is the answer to 5^{3x}=33(7^x) ?
The answer to 5^{3x}=33(7^x) is x=(ln(33))/(3ln(5)-ln(7))
Solve for x: 5^{3x}=33(7^x)
The solution is x=(ln(33))/(3ln(5)-ln(7))