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Exponential and Logarithmic Equations


Exponential and logarithmic functions can appear in exponential and logarithmic equations in Math.

These are often equations that we can solve just like we would solve a linear or quadratic equation for example.






Exponential and Logarithmic Equations Examples



Exponential Equations


Say we had the following.     2^x \space=\space 8.3

Where we want to find the  x  value.

We can take a  \tt{log}  of both sides to help us.

\tt{log}_{10}  or the natural log  \tt{ln}  are best as they appear are on a standard calculator.


So we proceed as follows.

2^x \space=\space 8.3       =>       {\tt{log}}_{10}\space2^x \space=\space {\tt{log}}_{10}\space8.3

x{\tt{log}}_{10}\space2 \space=\space {\tt{log}}_{10}\space8.3

x \space=\space {\Large{\frac{{\tt{log}}_{10}\space8.3}{{\tt{log}}_{10}\space2}}} \space\space = \space\space 3.053




Examples    



(1.1) 

a)     Solve   7^x \space=\space 23.8.

Solution   

7^x \space=\space 23.8       =>       {\tt{log}}_{10}\space7^x \space=\space {\tt{log}}_{10}\space23.8

x{\tt{log}}_{10}\space7 \space=\space {\tt{log}}_{10}\space23.8

x \space=\space {\Large{\frac{{\tt{log}}_{10}\space23.8}{{\tt{log}}_{10}\space2}}} \space\space = \space\space 1.6289


b)     Solve   7^x \space=\space 0.6.

Solution   

7^x \space=\space 0.6       =>       {\tt{log}}_{10}\space7^x \space=\space {\tt{log}}_{10}\space0.6

x{\tt{log}}_{10}\space7 \space=\space {\tt{log}}_{10}\space0.6

x \space=\space {\Large{\frac{{\tt{log}}_{10}\space0.6}{{\tt{log}}_{10}\space2}}} \space\space = \space\space {\text{-}}0.2625




(1.2) 

Solve   6^{2x} \space=\space 14.

Solution   

6^{2x} \space=\space 14       =>       {\tt{log}}_{10}\space6^{2x} \space=\space {\tt{log}}_{10}\space14

2x{\tt{log}}_{10}\space6 \space=\space {\tt{log}}_{10}\space14

2x \space=\space {\Large{\frac{{\tt{log}}_{10}\space14}{{\tt{log}}_{10}\space6}}} \space\space = \space\space {\text{-}}1.472886

x \space=\space {\Large{\frac{1.472886}{2}}} \space=\space 0.7364




(1.3) 

Solve   2^{x{\text{--}}3} \space=\space 12.

Solution   

2^{x{\text{--}}3} \space=\space 12       =>       {\tt{log}}_{10}\space2^{x{\text{--}}3} \space=\space {\tt{log}}_{10}\space12

(x\space{\text{--}}\space3){\tt{log}}_{10}\space2 \space=\space {\tt{log}}_{10}\space12

x\space{\text{--}}\space3 \space=\space {\Large{\frac{{\tt{log}}_{10}\space12}{{\tt{log}}_{10}\space2}}}

x \space=\space {\Large{\frac{{\tt{log}}_{10}\space12}{{\tt{log}}_{10}\space2}}} + 3 \space\space = \space\space 3.585 + 3 \space = \space 6.585




(1.4) 

Solve    {\Large{\frac{e^{3x\space{\text{--}}\space1}}{e^{2x\space{\text{--}}\space1}}}} \space=\space 8.

Solution   

This example may look a bit daunting at first.

But as  e  is the same base number in both the numerator and the denominator.
We can make use of the quotient rule.

Quotient Rule:   {\Large{\frac{a^b}{a^c}}} \space=\space a^{b \space{\text{--}}\space c}

e^{(3x\space{\text{--}}\space1)\space{\text{--}}\space(2x\space{\text{--}}\space1)} \space=\space 8

e^{3x\space{\text{--}}\space1 \space{\text{--}}\space 2x + 1} \space=\space 8       =>       e^x \space=\space 8

Now we can apply  \tt{ln}  to both sides, as this will undo  e.

{\tt{ln}}\space e^x \space=\space {\tt{ln}}\space8       =>       x{\tt{ln}}\space e \space=\space {\tt{ln}}\space8         ( {\tt{ln}}e = 1 )

x \space\space=\space\space {\tt{ln}}\space8 \space\space=\space\space 2.0794




(1.5) 

Solve    e^{2x} \space{\text{--}}\space 3 e^x + 10 \space=\space 0.

Solution   

e^{2x} \space{\text{--}}\space 3 e^x + 10 \space=\space 0     =>     (e^x)^2 \space{\text{--}}\space 3 e^x + 10 \space=\space 0

We can treat this as a quadratic equation and solve as such. It’s helpful to label e^x as a variable, say a.

a^2 \space{\text{--}}\space 3a + 10 \space=\space 0     =>     (a \space{\text{--}}\space 2)(a \space{\text{--}}\space 1) \space=\space 0

Now:

(e^x \space{\text{--}}\space 2)(e^x \space{\text{--}}\space 1) \space=\space 0

e^x = 2 \space \space , \space\space e^x = 1

We can apply  \tt{ln}  to both sides, as this will undo  e.

e^x = 2

{\tt{ln}}\space e^x = {\tt{ln}}\space2

x = {\tt{ln}}\space2

x = 0.69
e^x = 1

{\tt{ln}}\space e^x = {\tt{ln}}\space1

x = {\tt{ln}}\space1

x = 0







Logarithmic Equations Examples



(2.1) 

Solve for  x.       4\space {\tt{log}}_{10}(3x \space {\text{--}} \space 1) \space=\space 2

Solution   

{\tt{log}}_{10}(3x \space {\text{--}} \space 1) \space=\space {\Large{\frac{2}{4}}}       =>       {\tt{log}}_{10}(3x \space {\text{--}} \space 1) \space=\space {\Large{\frac{1}{2}}}

Now:

3x \space {\text{--}} \space 1 \space=\space 10^{\frac{1}{2}}       =>       3x \space {\text{--}} \space 1 \space=\space \sqrt{10}

3x \space=\space \sqrt{10} + 1     ,     3x \space=\space 4.16228

x \space=\space {\Large{\frac{4.16228}{3}}}     ,     x \space=\space 1.3874




(2.2) 

Solve for  x.       {\tt{log}}_3\space x + {\tt{\log}}_3\space5 \space=\space {\tt{log}}_3\space20

Solution   

We can make use of the log rule,   {\tt{log}}_a (mn) \space {\small{=}} \space {\tt{log}}_a (m) + {\tt{log}}_a (n).

{\tt{log}}_3\space 5x \space=\space {\tt{log}}_3\space20

5x \space=\space 20       ,       x \space=\space 4




(2.3) 

Solve for  x.       3 \space{\tt{ln}}\space4 + {\tt{ln}}(2x) \space=\space {\tt{ln}}\space128

Solution   

3 \space{\tt{ln}}\space4 \space=\space {\tt{ln}}\space128 \space{\text{--}}\space {\tt{ln}}(2x)

3 \space{\tt{ln}}\space4 \space=\space {\tt{ln}}\space({\Large{\frac{128}{2x}}})       =>       {\tt{ln}}\space4^3 \space=\space {\tt{ln}}\space({\Large{\frac{128}{2x}}})

{\tt{ln}}\space64 \space=\space {\tt{ln}}\space({\Large{\frac{128}{2x}}})

64 \space=\space {\Large{\frac{128}{2x}}}       =>       x \space=\space 1





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