Logarithm Equation Calculator

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Calculators

Algebra

Logarithm Equation Solver

\[ \log_{b}x = y \]

Enter any two values or fractions
to solve for the third value

b =

x =

y =

Answer:

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Calculator Use

This calculator will solve the basic log equation log_bx = y for any one of the variables as long as you enter the other two.

The logarithmic equation is solved using the logarithmic function:

\( x = \log_{b}b^x \)

which is equivalently

\( x = b^{log_{b}x} \)

How to solve the logarithmic equation

If we have the equation used in the Logarithm Equation Calculator

\( \log_{b}x = y \;(1) \)

We can say the following is also true

\( b^{\log_{b}x} = b^{y} \;(2) \)

Using the logarithmic function where

\( x = b^{log_{b}x} \)

We can rewrite our equation (2) to solve for x

\( x = b^{y} \;(3) \)

Solving for b in equation (3) we have

\( b = \sqrt[y]{x} \)

Solving for y in equation (3)

\( x = b^{y} \;(3) \)

take the log of both sides:

\( \log_{10}x = \log_{10}b^y \)

Using logarithmic identity we rewrite the equation:

\( \log_{10}x = y \cdot \log_{10}b \)

Dividing both sides by log b:

\( y = \dfrac{\log_{10}x}{\log_{10}b} = \dfrac{\log_{}x}{\log_{}b} \)

Note that writing log without the subscript for the base it is assumed to be log base 10 as in log₁₀.

Example 1: Solve for y in the following logarithmic equation

If we have

\( \log_{3}5 = y \)

then it is also true that

\( 3^{\log_{3}5} = 3^{y} \)

Using the logarithmic function we can rewrite the left side of the equation and we get

\( 5 = 3^{y} \)

To solve for y, first take the log of both sides:

\( \log_{}5 = \log_{}3^y \)

By the identity log x^y = y · log x we get:

\( \log_{}5 = y \cdot \log_{}3 \)

Dividing both sides by log 3:

\( y = \dfrac{\log_{}5}{\log_{}3} \)

Using a calculator we can find that log 5 ≈ 0.69897 and log 3 ≈ 0.4771 2 then our equation becomes:

\( n = \dfrac{\log_{}5}{\log_{}3} = \dfrac{0.69897}{0.47712} = 1.46497 \)

Therefore, putting y back into our original equation

\( \log_{3}5 = 1.46497 \)

Example 2: Solve for b in the following logarithmic equation

If we have

\( \log_{b}16 = 2 \)

then it is also true that

\( b^{\log_{b}16} = b^{2} \)

Using the logarithmic function we can rewrite the left side of the equation and we get

\( 16 = b^{2} \)

Solving for b by taking the 2nd root of both sides of the equation

\( b = \sqrt[2]{16} = 4 \)

Therefore, putting b back into our original equation

\( \log_{4}16 = 2 \)