What is the answer to solvefor x,log_{x+1}(y)=2 ?

The answer to solvefor x,log_{x+1}(y)=2 is x=e^{(ln(y))/2}-1 {0 1}

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solvefor x,log_{x+1}(y)=2

solve for

x, lo g_{x + 1} (y) = 2

x = e^{\frac{l n ( y )}{2}} - 1 {0 < y < 1 or y > 1}

Solution steps

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

Apply log rules

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

Multiply both sides by

ln (x + 1)

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

Simplify

ln (y) = 2 ln (x + 1)

Switch sides

2 ln (x + 1) = ln (y)

Divide both sides by

2

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

Apply log rules

x + 1 = e^{\frac{l n ( y )}{2}}

Solve

x + 1 = e^{\frac{l n ( y )}{2}} : x = e^{\frac{l n ( y )}{2}} - 1

x = e^{\frac{l n ( y )}{2}} - 1

Verify Solutions:

x = e^{\frac{l n ( y )}{2}} - 1 {0 < y < 1 or y > 1}

The solution is

x = e^{\frac{l n ( y )}{2}} - 1 {0 < y < 1 or y > 1}

\frac{1}{x} = \frac{ln ( x ) + x}{x}

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

9.1 = 10 lo g_{10} (\frac{4.8}{x})

t, lo g_{5} (t) = u

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

What is the answer to solvefor x,log_{x+1}(y)=2 ?
The answer to solvefor x,log_{x+1}(y)=2 is x=e^{(ln(y))/2}-1 {0<y<1\lor y>1}