What is the answer to log_{x+1}(5x-1)=2 ?

The answer to log_{x+1}(5x-1)=2 is x=2,x=1

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log_{x+1}(5x-1)=2

lo g_{x + 1} (5 x - 1) = 2

x = 2, x = 1

Solution steps

lo g_{x + 1} (5 x - 1) = 2

Apply log rules

\frac{ln ( 5 x - 1 )}{ln ( x + 1 )} = 2

Multiply both sides by

ln (x + 1)

\frac{ln ( 5 x - 1 )}{ln ( x + 1 )} ln (x + 1) = 2 ln (x + 1)

Simplify

ln (5 x - 1) = 2 ln (x + 1)

Apply log rules

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

Solve

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

x = 2, x = 1

Verify Solutions:

x = 2

True

, x = 1

True

The solutions are

x = 2, x = 1

lo g_{4} (x + 12) = 2

lo g_{10} ((x - 1)^{2}) = 0

lo g_{- 2 x - 4} (2 x^{2} - 14) = 2

x, 3.5 = 3.75 + lo g_{10} (x)

lo g_{\frac{1}{3}} (x) = \frac{1}{27}

What is the answer to log_{x+1}(5x-1)=2 ?
The answer to log_{x+1}(5x-1)=2 is x=2,x=1
Solve for x: log_{x+1}(5x-1)=2
The solution is x=2,x=1