What is the answer to 2*log_{10}(x)=log_{10}(2)+log_{10}(3x-4) ?

The answer to 2*log_{10}(x)=log_{10}(2)+log_{10}(3x-4) is x=4,x=2

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2*log_{10}(x)=log_{10}(2)+log_{10}(3x-4)

2 \cdot lo g_{10} (x) = lo g_{10} (2) + lo g_{10} (3 x - 4)

x = 4, x = 2

Solution steps

2 lo g_{10} (x) = lo g_{10} (2) + lo g_{10} (3 x - 4)

Apply log rules

x^{2} = 2 (3 x - 4)

Solve

x^{2} = 2 (3 x - 4) : x = 4, x = 2

x = 4, x = 2

Verify Solutions:

x = 4

True

, x = 2

True

The solutions are

x = 4, x = 2

ln (r) - ln (b) = 2 ln (a) - ln (c)

lo g_{10} (x + 2) - lo g_{10} (x) = 1

0.051 = lo g_{10} (x)

- lo g_{10} (b + 98) = 0.125

x, 2 y^{2} + 3 y - 5 x = ln (x + y^{2}) - 1

What is the answer to 2*log_{10}(x)=log_{10}(2)+log_{10}(3x-4) ?
The answer to 2*log_{10}(x)=log_{10}(2)+log_{10}(3x-4) is x=4,x=2
Solve for x: 2*log_{10}(x)=log_{10}(2)+log_{10}(3x-4)
The solution is x=4,x=2