What is the answer to log_{4}(x)+6log_{x}(4)-5=0 ?

The answer to log_{4}(x)+6log_{x}(4)-5=0 is x=64,x=16

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log_{4}(x)+6log_{x}(4)-5=0

lo g_{4} (x) + 6 lo g_{x} (4) - 5 = 0

x = 64, x = 16

Solution steps

lo g_{4} (x) + 6 lo g_{x} (4) - 5 = 0

Apply log rules

lo g_{4} (x) + \frac{6}{lo g _{4} ( x )} - 5 = 0

Rewrite the equation with

lo g_{4} (x) = u

u + \frac{6}{u} - 5 = 0

Solve

u + \frac{6}{u} - 5 = 0 : u = 3, u = 2

u = 3, u = 2

Substitute back

u = lo g_{4} (x),

solve for

x

Solve

lo g_{4} (x) = 3 : x = 64

Solve

lo g_{4} (x) = 2 : x = 16

x = 64, x = 16

Verify Solutions:

x = 64

True

, x = 16

True

The solutions are

x = 64, x = 16

7.4 = 6.1 + lo g_{10} (x)

2 ln (x) = ln (x + 6)

lo g_{x} (x^{2} - 5 x - 4) = 1

ln (4 - c) = 0

lo g_{10} (7 x + 3) = 2

What is the answer to log_{4}(x)+6log_{x}(4)-5=0 ?
The answer to log_{4}(x)+6log_{x}(4)-5=0 is x=64,x=16
Solve for x: log_{4}(x)+6log_{x}(4)-5=0
The solution is x=64,x=16