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log 2 ( 2 ) + log 2 ( 3 x 5 ) = 3             log 2 ( 2 ( 3 x 5 ) ) = 3 Apply the product rule of logarithms .                log 2 ( 6 x 10 ) = 3 Distribute .                                 2 3 = 6 x 10 Apply the definition of a logarithm .                                    8 = 6 x 10 Calculate  2 3 .                                  18 = 6 x Add 10 to both sides .                                   x = 3 Divide by 6 .

Using the definition of a logarithm to solve logarithmic equations

For any algebraic expression S and real numbers b and c , where b > 0 ,   b 1 ,

log b ( S ) = c if and only if b c = S

Using algebra to solve a logarithmic equation

Solve 2 ln x + 3 = 7.

2 ln x + 3 = 7       2 ln x = 4 Subtract 3 .         ln x = 2 Divide by 2 .             x = e 2 Rewrite in exponential form .
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Solve 6 + ln x = 10.

x = e 4

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Using algebra before and after using the definition of the natural logarithm

Solve 2 ln ( 6 x ) = 7.

2 ln ( 6 x ) = 7    ln ( 6 x ) = 7 2 Divide by 2 .          6 x = e ( 7 2 ) Use the definition of  ln .            x = 1 6 e ( 7 2 ) Divide by 6 .
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Solve 2 ln ( x + 1 ) = 10.

x = e 5 1

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Using a graph to understand the solution to a logarithmic equation

Solve ln x = 3.

ln x = 3 x = e 3 Use the definition of the natural logarithm .

[link] represents the graph of the equation. On the graph, the x -coordinate of the point at which the two graphs intersect is close to 20. In other words e 3 20. A calculator gives a better approximation: e 3 20.0855.

Graph of two questions, y=3 and y=ln(x), which intersect at the point (e^3, 3) which is approximately (20.0855, 3).
The graphs of y = ln x and y = 3 cross at the point (e 3 , 3 ) , which is approximately (20.0855, 3).
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Use a graphing calculator to estimate the approximate solution to the logarithmic equation 2 x = 1000 to 2 decimal places.

x 9.97

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Using the one-to-one property of logarithms to solve logarithmic equations

As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers x > 0 , S > 0 , T > 0 and any positive real number b , where b 1 ,

log b S = log b T  if and only if  S = T .

For example,

If   log 2 ( x 1 ) = log 2 ( 8 ) , then  x 1 = 8.

So, if x 1 = 8 , then we can solve for x , and we get x = 9. To check, we can substitute x = 9 into the original equation: log 2 ( 9 1 ) = log 2 ( 8 ) = 3. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.

For example, consider the equation log ( 3 x 2 ) log ( 2 ) = log ( x + 4 ) . To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for x :

log ( 3 x 2 ) log ( 2 ) = log ( x + 4 )              log ( 3 x 2 2 ) = log ( x + 4 ) Apply the quotient rule of logarithms .                      3 x 2 2 = x + 4 Apply the one to one property of a logarithm .                      3 x 2 = 2 x + 8 Multiply both sides of the equation by  2.                               x = 10 Subtract 2 x  and add 2 .

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Practice Key Terms 1

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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