What is the answer to log_{2}(x+1)=1+6log_{x+1}(2) ?

The answer to log_{2}(x+1)=1+6log_{x+1}(2) is x=7,x=-3/4

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log_{2}(x+1)=1+6log_{x+1}(2)

lo g_{2} (x + 1) = 1 + 6 lo g_{x + 1} (2)

x = 7, x = - \frac{3}{4}

Solution steps

lo g_{2} (x + 1) = 1 + 6 lo g_{x + 1} (2)

Apply log rules

lo g_{2} (x + 1) = 1 + \frac{6}{lo g _{2} ( x + 1 )}

Rewrite the equation with

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

u = 1 + \frac{6}{u}

Solve

u = 1 + \frac{6}{u} : u = 3, u = - 2

u = 3, u = - 2

Substitute back

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

solve for

x

Solve

lo g_{2} (x + 1) = 3 : x = 7

Solve

lo g_{2} (x + 1) = - 2 : x = - \frac{3}{4}

x = 7, x = - \frac{3}{4}

Verify Solutions:

x = 7

True

, x = - \frac{3}{4}

True

The solutions are

x = 7, x = - \frac{3}{4}

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

- lo g_{10} (x) = 1.55

2 lo g_{3} (x) = 3 lo g_{3} (5)

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

ln (2 x) = 4 ln (4 x)

What is the answer to log_{2}(x+1)=1+6log_{x+1}(2) ?
The answer to log_{2}(x+1)=1+6log_{x+1}(2) is x=7,x=-3/4
Solve for x: log_{2}(x+1)=1+6log_{x+1}(2)
The solution is x=7,x=-3/4