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

lo g_{4} (x^{2} - 25) = 1 + lo g_{4} (x - 1)

lo g_{10} (7 x) = - 2

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

ln (2 x - 3) = 2

lo g_{10} (9 x) = - 2

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