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

The answer to log_{x}(5x)-log_{6}(4)=0 is x= 1/(e^{\frac{ln(5)){1-2log_{6}(2)}}}

Solve for x: log_{x}(5x)-log_{6}(4)=0

The solution is x= 1/(e^{\frac{ln(5)){1-2log_{6}(2)}}}

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

x = \frac{1}{e ^{\frac{l n ( 5 )}{1 - 2 l o g _{6} ( 2 )}}}

Solution steps

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

Move

lo g_{6} (4)

to the right side

lo g_{x} (5 x) = lo g_{6} (4)

Apply log rules

\frac{ln ( 5 x )}{ln ( x )} = lo g_{6} (4)

Multiply both sides by

ln (x)

\frac{ln ( 5 x )}{ln ( x )} ln (x) = lo g_{6} (4) ln (x)

Apply log rules

x = \frac{1}{e ^{\frac{l n ( 5 )}{1 - 2 l o g _{6} ( 2 )}}}

Verify Solutions:

x = \frac{1}{e ^{\frac{l n ( 5 )}{1 - 2 l o g _{6} ( 2 )}}}

True

The solution is

x = \frac{1}{e ^{\frac{l n ( 5 )}{1 - 2 l o g _{6} ( 2 )}}}

lo g_{2} (x + 5) = 6 - x

lo g_{5} (y^{2} + 17) = 4

0.6 = ln (y)

lo g_{2 x - 5} (3 x^{2} - 7 x - 11) = 2

ln (x^{0.5}) = (ln (x))^{0.5}

What is the answer to log_{x}(5x)-log_{6}(4)=0 ?
The answer to log_{x}(5x)-log_{6}(4)=0 is x= 1/(e^{\frac{ln(5)){1-2log_{6}(2)}}}
Solve for x: log_{x}(5x)-log_{6}(4)=0
The solution is x= 1/(e^{\frac{ln(5)){1-2log_{6}(2)}}}