What is the answer to (log_{10}(2x^2+1))/(log_{10)(x+1)}=2 ?

The answer to (log_{10}(2x^2+1))/(log_{10)(x+1)}=2 is x=2

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(log_{10}(2x^2+1))/(log_{10)(x+1)}=2

\frac{lo g _{10} ( 2 x ^{2} + 1 )}{lo g _{10} ( x + 1 )} = 2

x = 2

Solution steps

\frac{lo g _{10} ( 2 x ^{2} + 1 )}{lo g _{10} ( x + 1 )} = 2

Multiply both sides by

lo g_{10} (x + 1)

\frac{lo g _{10} ( 2 x ^{2} + 1 )}{lo g _{10} ( x + 1 )} lo g_{10} (x + 1) = 2 lo g_{10} (x + 1)

Simplify

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

Apply log rules

2 x^{2} + 1 = (x + 1)^{2}

Solve

2 x^{2} + 1 = (x + 1)^{2} : x = 0, x = 2

x = 0, x = 2

Verify Solutions:

x = 0

False

, x = 2

True

The solution is

x = 2

40 = 20 \cdot lo g_{10} (\frac{x}{0.0002})

lo g_{4} (x) (x - 2) = lo g_{2} (x)

lo g_{10} (c) = 0

lo g_{3} (x + 2) + lo g_{3} (x + 1) = lo g_{3} (x^{2} + 1)

lo g_{22} (2 x) - 5 lo g_{2} (x) = 0

What is the answer to (log_{10}(2x^2+1))/(log_{10)(x+1)}=2 ?
The answer to (log_{10}(2x^2+1))/(log_{10)(x+1)}=2 is x=2
Solve for x: (log_{10}(2x^2+1))/(log_{10)(x+1)}=2
The solution is x=2