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

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

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

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

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

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

Decimal Notation

x = 1.5

Solution steps

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

Apply log rules

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

Multiply both sides by

ln (x)

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

Simplify

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

Switch sides

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

Divide both sides by

- 4

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

Apply log rules

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

Verify Solutions:

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

True

The solution is

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

2 \cdot ln (x) - ln (2) = ln (2 x - 1)

lo g_{3} (4 x - 5) = lo g_{3} (x + 3)

20 lo g_{10} (x) = 29.5

lo g_{3} (x + 58) = lo g_{3} (x + 2) + 2

lo g_{6} (x + 5) = 36

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