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

The answer to log_{10}(x)=1+log_{10}(3x-4) is x= 40/29

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

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

x = \frac{40}{29}

Decimal Notation

x = 1.37931 \dots

Solution steps

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

Apply log rules

x = 10 (3 x - 4)

Solve

x = 10 (3 x - 4) : x = \frac{40}{29}

x = \frac{40}{29}

Verify Solutions:

x = \frac{40}{29}

True

The solution is

x = \frac{40}{29}

lo g_{\frac{1}{2}} (5 (x - 1)) = 3

lo g_{2} (x) - lo g_{2} (x + 1) = 2

lo g_{2} (x + 5) = 2 + lo g_{2} (x + 8)

2 lo g_{3} (x) = 2 + lo g_{3} (x - 2)

ln (4 + a e^{b t}) = 0

What is the answer to log_{10}(x)=1+log_{10}(3x-4) ?
The answer to log_{10}(x)=1+log_{10}(3x-4) is x= 40/29
Solve for x: log_{10}(x)=1+log_{10}(3x-4)
The solution is x= 40/29