What is the answer to log_{10}(3)(4x-3)+log_{10}(3)(4x-1)=1 ?

The answer to log_{10}(3)(4x-3)+log_{10}(3)(4x-1)=1 is x=(1+4log_{10}(3))/(8log_{10)(3)}

Solve for x: log_{10}(3)(4x-3)+log_{10}(3)(4x-1)=1

The solution is x=(1+4log_{10}(3))/(8log_{10)(3)}

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lo g_{10} (3) (4 x - 3) + lo g_{10} (3) (4 x - 1) = 1

x = \frac{1 + 4 lo g _{10} ( 3 )}{8 lo g _{10} ( 3 )}

Solution steps

lo g_{10} (3) (4 x - 3) + lo g_{10} (3) (4 x - 1) = 1

Expand

lo g_{10} (3) (4 x - 3) + lo g_{10} (3) (4 x - 1) : 8 lo g_{10} (3) x - 4 lo g_{10} (3)

8 lo g_{10} (3) x - 4 lo g_{10} (3) = 1

Move

4 lo g_{10} (3)

to the right side

8 lo g_{10} (3) x = 1 + 4 lo g_{10} (3)

Divide both sides by

8 lo g_{10} (3)

x = \frac{1 + 4 lo g _{10} ( 3 )}{8 lo g _{10} ( 3 )}

15 + 1 - 12 q = - 18 q - 14

- \frac{2}{3} = - \frac{4}{5} x - \frac{1}{4}

\frac{2 x - 1}{5} = \frac{x + 1}{4}

y = - 2^{3}

\frac{1}{4} x - 3 = 1

What is the answer to log_{10}(3)(4x-3)+log_{10}(3)(4x-1)=1 ?
The answer to log_{10}(3)(4x-3)+log_{10}(3)(4x-1)=1 is x=(1+4log_{10}(3))/(8log_{10)(3)}
Solve for x: log_{10}(3)(4x-3)+log_{10}(3)(4x-1)=1
The solution is x=(1+4log_{10}(3))/(8log_{10)(3)}