What is the answer to 12log_{x}(5)+log_{5}(x)-7=0 ?

The answer to 12log_{x}(5)+log_{5}(x)-7=0 is x=625,x=125

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12log_{x}(5)+log_{5}(x)-7=0

12 lo g_{x} (5) + lo g_{5} (x) - 7 = 0

x = 625, x = 125

Solution steps

12 lo g_{x} (5) + lo g_{5} (x) - 7 = 0

Apply log rules

\frac{12}{lo g _{5} ( x )} + lo g_{5} (x) - 7 = 0

Rewrite the equation with

lo g_{5} (x) = u

\frac{12}{u} + u - 7 = 0

Solve

\frac{12}{u} + u - 7 = 0 : u = 4, u = 3

u = 4, u = 3

Substitute back

u = lo g_{5} (x),

solve for

x

Solve

lo g_{5} (x) = 4 : x = 625

Solve

lo g_{5} (x) = 3 : x = 125

x = 625, x = 125

Verify Solutions:

x = 625

True

, x = 125

True

The solutions are

x = 625, x = 125

4 ln (x + 1) = 12

x, g = lo g_{4} (x - 3)

lo g_{10} (2 x + 3) = lo g_{10} (x + 5)

ln (x^{2} - x + 17) = \frac{3}{4}

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

What is the answer to 12log_{x}(5)+log_{5}(x)-7=0 ?
The answer to 12log_{x}(5)+log_{5}(x)-7=0 is x=625,x=125
Solve for x: 12log_{x}(5)+log_{5}(x)-7=0
The solution is x=625,x=125