What is the answer to log_{x}(81/16)=-4 ?

The answer to log_{x}(81/16)=-4 is x= 1/(e^{\frac{ln(81/16)){4}}}

Solve for x: log_{x}(81/16)=-4

The solution is x= 1/(e^{\frac{ln(81/16)){4}}}

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lo g_{x} (\frac{81}{16}) = - 4

x = \frac{1}{e ^{\frac{l n ( \frac{81}{16} )}{4}}}

Decimal Notation

x = 0.66666 \dots

Solution steps

lo g_{x} (\frac{81}{16}) = - 4

Apply log rules

\frac{ln ( \frac{81}{16} )}{ln ( x )} = - 4

Multiply both sides by

ln (x)

\frac{ln ( \frac{81}{16} )}{ln ( x )} ln (x) = - 4 ln (x)

Simplify

ln (\frac{81}{16}) = - 4 ln (x)

Switch sides

- 4 ln (x) = ln (\frac{81}{16})

Divide both sides by

- 4

ln (x) = - \frac{ln ( \frac{81}{16} )}{4}

Apply log rules

x = \frac{1}{e ^{\frac{l n ( \frac{81}{16} )}{4}}}

Verify Solutions:

x = \frac{1}{e ^{\frac{l n ( \frac{81}{16} )}{4}}}

True

The solution is

x = \frac{1}{e ^{\frac{l n ( \frac{81}{16} )}{4}}}

lo g_{6} (x + 4) + lo g_{6} (x + 3) = 1

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

lo g_{x} (\frac{1}{81}) = - 4

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

2 x ln (x) - x = 0

What is the answer to log_{x}(81/16)=-4 ?
The answer to log_{x}(81/16)=-4 is x= 1/(e^{\frac{ln(81/16)){4}}}
Solve for x: log_{x}(81/16)=-4
The solution is x= 1/(e^{\frac{ln(81/16)){4}}}