What is the answer to log_{n}(10^{-9})=20 ?

The answer to log_{n}(10^{-9})=20 is n= 1/(e^{\frac{9ln(10)){20}}}

Solve for n: log_{n}(10^{-9})=20

The solution is n= 1/(e^{\frac{9ln(10)){20}}}

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lo g_{n} (1 0^{- 9}) = 20

n = \frac{1}{e ^{\frac{9 l n ( 10 )}{20}}}

Decimal Notation

n = 0.35481 \dots

Solution steps

lo g_{n} (1 0^{- 9}) = 20

Apply log rules

\frac{ln ( 1 0 ^{- 9} )}{ln ( n )} = 20

Multiply both sides by

ln (n)

\frac{ln ( 1 0 ^{- 9} )}{ln ( n )} ln (n) = 20 ln (n)

Apply log rules

n = \frac{1}{e ^{\frac{9 l n ( 10 )}{20}}}

Verify Solutions:

n = \frac{1}{e ^{\frac{9 l n ( 10 )}{20}}}

True

The solution is

n = \frac{1}{e ^{\frac{9 l n ( 10 )}{20}}}

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

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

5 ln (4 x - 9) + 7 = - 8

ln (n^{2} + 2 n + 1) = 0

3 ln (2 x) = - 3

What is the answer to log_{n}(10^{-9})=20 ?
The answer to log_{n}(10^{-9})=20 is n= 1/(e^{\frac{9ln(10)){20}}}
Solve for n: log_{n}(10^{-9})=20
The solution is n= 1/(e^{\frac{9ln(10)){20}}}