# 6.6 Exponential and logarithmic equations  (Page 4/8)

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## Using the definition of a logarithm to solve logarithmic equations

For any algebraic expression $\text{\hspace{0.17em}}S\text{\hspace{0.17em}}$ and real numbers $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}c,$ where

## Using algebra to solve a logarithmic equation

Solve $\text{\hspace{0.17em}}2\mathrm{ln}x+3=7.$

Solve $\text{\hspace{0.17em}}6+\mathrm{ln}x=10.$

$x={e}^{4}$

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

Solve $\text{\hspace{0.17em}}2\mathrm{ln}\left(6x\right)=7.$

Solve $\text{\hspace{0.17em}}2\mathrm{ln}\left(x+1\right)=10.$

$x={e}^{5}-1$

## Using a graph to understand the solution to a logarithmic equation

Solve $\text{\hspace{0.17em}}\mathrm{ln}x=3.$

[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 $\text{\hspace{0.17em}}{e}^{3}\approx 20.\text{\hspace{0.17em}}$ A calculator gives a better approximation: $\text{\hspace{0.17em}}{e}^{3}\approx 20.0855.$

Use a graphing calculator to estimate the approximate solution to the logarithmic equation $\text{\hspace{0.17em}}{2}^{x}=1000\text{\hspace{0.17em}}$ to 2 decimal places.

$x\approx 9.97$

## 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 $\text{\hspace{0.17em}}x>0,$ $S>0,$ $T>0\text{\hspace{0.17em}}$ and any positive real number $\text{\hspace{0.17em}}b,$ where $\text{\hspace{0.17em}}b\ne 1,$

For example,

So, if $\text{\hspace{0.17em}}x-1=8,$ then we can solve for $\text{\hspace{0.17em}}x,$ and we get $\text{\hspace{0.17em}}x=9.\text{\hspace{0.17em}}$ To check, we can substitute $\text{\hspace{0.17em}}x=9\text{\hspace{0.17em}}$ into the original equation: $\text{\hspace{0.17em}}{\mathrm{log}}_{2}\left(9-1\right)={\mathrm{log}}_{2}\left(8\right)=3.\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}\mathrm{log}\left(3x-2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(x+4\right).\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}x:$

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why two x + seven is equal to nineteen.
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2x + 7 =19
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2x +7=19. 2x=19 - 7 2x=12 x=6
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because x is 6
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what is the value of x in 4x-2+3
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if 4x-2+3 = 0 then 4x = 2-3 4x = -1 x = -(1÷4) is the answer.
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