What is the answer to log_{10}(4^{2x})=log_{10}(9) ?

The answer to log_{10}(4^{2x})=log_{10}(9) is x=(ln(3))/(2ln(2))

Solve for x: log_{10}(4^{2x})=log_{10}(9)

The solution is x=(ln(3))/(2ln(2))

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lo g_{10} (4^{2 x}) = lo g_{10} (9)

x = \frac{ln ( 3 )}{2 ln ( 2 )}

Solution steps

lo g_{10} (4^{2 x}) = lo g_{10} (9)

Rewrite the equation with

4^{2 x} = u

lo g_{10} (u) = lo g_{10} (9)

Solve

lo g_{10} (u) = lo g_{10} (9) : u = 9

u = 9

Substitute back

u = 4^{2 x},

solve for

x

Solve

4^{2 x} = 9 : x = \frac{ln ( 3 )}{2 ln ( 2 )}

x = \frac{ln ( 3 )}{2 ln ( 2 )}

67 e^{\frac{x}{5}} = 51

- 15 (1 - 2^{x}) = 123933

2^{x} = (0.0625)^{- 14}

2 7^{2 x - 3} = 93

2^{x} = (0.25)^{x - 1}

What is the answer to log_{10}(4^{2x})=log_{10}(9) ?
The answer to log_{10}(4^{2x})=log_{10}(9) is x=(ln(3))/(2ln(2))
Solve for x: log_{10}(4^{2x})=log_{10}(9)
The solution is x=(ln(3))/(2ln(2))