Solving Exponential and Logarithmic Equations
Learning Objective(s)
· Solve exponential equations.
· Solve logarithmic equations.
Introduction
As you know, algebra often requires you to solve equations to find unknown values. This is also true for exponential and logarithmic equations. There are some strategies that you can use, along with some properties you’ve learned, that you can use to solve those equations.
You may be able to look at an equation like 4^{x}^{ }= 16 and solve it by asking yourself, “4 to what power is 16? 4^{2} is 16, so x = 2.” Equations like 4^{x} = 17 are a bit more difficult. You know x must be a little more than 2, because 17 is just a little more than 16. One way to find x with more precision, though, is by using logarithms.
When you have solved other algebraic equations, you often relied on the idea that you can change both sides of the equation in the same way and still get a true equation. This is true with logarithms, too: If x = y, then log_{b}x = log_{b}y, no matter what b is.
Let’s look at this with an equation whose solution you already know: 4^{x} = 16. You can use either the common log or the natural log. In the following example, you will use the common log.
Example  
Problem  Solve 4^{x} = 16.  
 4^{x} = 16 log 4^{x} = log 16
 Take the common logarithm of both sides. (Remember, when no base is written, that means the base is 10.)
What can you do with that new equation? 
 log 4^{x} = log 16 x log 4 = log 16  Use the power property of logarithms to simplify the logarithm on the left side of the equation. 
 x log 4 = log 16
 Remember that log 4 is a number. You can divide both sides of the equation by log 4 to get x by itself. 
Answer 
 Use a calculator to evaluate the logarithms and the quotient. 
Just as you knew, x = 2. Now let’s try it with our more difficult example, 4^{x} = 17. The procedure is exactly the same.
Example  
Problem  Solve 4^{x} = 17.  
 4^{x} = 17 log 4^{x} = log 17
 Take the common logarithm of both sides. 
 log 4^{x} = log 17 x log 4 = log 17  Use the power property of logarithms to simplify the logarithm on the left. 
 x log 4 = log 17
 Divide both sides by log 4 to get x by itself. 
Answer 
 Use a calculator to evaluate the logarithms and the quotient. 
You could have used either the common log or the natural log with the example above. You use one of these two bases, as you can then use your calculator to find the values.
Example  
Problem  Solve e^{2x} = 54.  
 e^{2x} = 54 ln e^{2x} = ln 54
 Since the base is e, use the natural logarithm. (If the base were 10, using common logarithms would be better.) 
 ln e^{2x} = ln 54 2x = ln 54  Remember that logarithms and exponential functions are inverses. When you have log_{b}b^{m}, the logarithm undoes the exponent, and the result is just m. So ln e^{2x} = log_{e}e^{2x} = 2x. 
 x =  Divide both sides by 2 to get x by itself. 
Answer  x = 1.99449...  Use a calculator to evaluate the logarithm and quotient on the right and you’re done! 
Another kind of exponential equation has exponential expressions on both sides. When the bases are the same, or the exponents are the same, you can just compare the parts that are different. Look at these examples.
Example  
Problem  Solve 3^{2x + 5 }= 3^{3x – 2}.  
 3^{2x + 5} = 3^{3x – 2}
 Here are two exponential expressions with the same base. If the two expressions are equal, then their exponents must be equal. (Think about that—if you have 3^{a} and 3^{b}, and a ≠ b, then 3^{a} can’t have the same value as 3^{b}.) 
 2x + 5 = 3x – 2  So, write a new equation that sets the exponents equal to each other. 
 5 = x – 2 7 = x  Solve the linear equation as you normally would. 
Check  3^{2(7) + 5} = 3^{3(7) – 2} 3^{19} = 3^{19}  Test the solution in the original equation.
No need to find 3^{19}. When both sides say the same thing, you know it’s correct! 
Answer  x = 7 

Example  
Problem  Solve (x + 4)^{8} = 7^{8}.  
 (x + 4)^{8} = 7^{8}
 Again, you have two exponential expressions that are equal to each other. In this case, both sides have the same exponent, and this means the bases must be equal. 
 x + 4 = 7  Write a new equation that sets the bases equal to each other. 
 x = 3  Solve the linear equation as you normally would. 
Check  (3 + 4)^{8} = 7^{8} 7^{8} = 7^{8} ^{ }  Test the solution in the original equation.
No need to find 7^{8}. When both sides say the same thing, you know it’s correct! 
Answer  x = 3 

Solve 10^{3x – 2} = 13.
A) x = 5 B) x = 1.03798… C) x = 1.52164… D) x = 3.11394…

There are several strategies you can use to solve logarithmic equations. The first is one you have used before: Rewrite the logarithmic equation as an exponential equation!
Example  
Problem  Solve ln x = 4.657. Give x to the thousandths place.  
 ln x = 4.657 log_{e} x = 4.657 e^{4.657} = x^{ }  Remember that natural logarithms have a base of e. Rewrite this logarithm as an exponential equation. 
Answer  105.3196… = x x ≈ 105.320  Use a calculator to evaluate e^{4.657}, and round to the nearest thousandth. 
This works regardless of the base.
Example  
Problem  Solve log_{7 }x = 3.843. Give x to the thousandths place.  
 log_{7} x = 3.843 7^{3.843} = x^{ }  Rewrite this logarithm as an exponential equation. 
Answer  1768.9345…= x x ≈ 1768.935  Use a calculator to evaluate 7^{3.843} and round to the nearest thousandth. 
Logarithmic equations may also involve inputs where the variable has a coefficient other than 1, or where the variable itself is squared. In these cases, you need to complete a few more steps in solving for the variable.
Example  
Problem  Solve log_{5}3x^{2} = 1.96. Give x to the hundredths place.  
 5^{1.96} = 3x^{2 }  Rewrite this logarithmic equation as an exponential equation. 
 23.44127… = 3x^{2}  Evaluate 5^{1.96}. 
 7.81375… = x^{2} x = ±2.7953… x ≈ ±2.80  Solve as you normally would. In this case, divide both sides by 3, then use the square root property to find the possible values for x. Don’t forget that when using the square root property, both positive and negative roots must be considered. Round to the nearest hundredth. 
Check  log_{5 }3x^{2} = 1.96 log_{5 }3(2.80)^{2} = 1.96 log_{5 }3(7.84) = 1.96 log_{5 }23.52 = 1.96
 Check your answer by substituting the value of x into the original equation. Because (2.80)^{2} and (+2.80)^{2} are both positive we don’t need to check +2.80 separately. 

 Apply the change of base formula to switch from base 5 to base 10.
The check shows that with rounding accounted for, a true statement results, so you know that the answer is correct. 
Answer  x ≈ ±2.80 

The equations may also include more than one logarithm. You can use the properties of logarithms to combine these logarithms into one logarithm. Note: You’ll find it helpful to record which properties you use at each step, both to help you be sure you’re using them properly and as a way to help you find errors.
Example  
Problem  Solve .  

 First notice that all of the logarithms have the same base. (These are common logarithms, so the bases are all 10.) When using the properties, it is absolutely necessary that the bases are the same.
 

 Use the power property to rewrite 2log 3 as log 3^{2} and to rewrite as .  

log 9 + log 4 – log 3 = log x  Evaluate the exponents.  
 log (9 • 4) – log 3 = log x log 36 – log 3 = log x  Use the product property, , to combine log 9 + log 4.  

log = log x log 12 = log x  Use the quotient property, , to combine log 36 – log 3.  
Answer  x = 12  Since the logarithm of 12 and the logarithm of x are equal, x must equal 12.  
Solve log x + log 3 = log 24.
A) 0.460… B) 2.892… C) 8 D) 21

Summary
There are several strategies that can be used to solve equations involving exponents and logarithms. Taking logarithms of both sides is helpful with exponential equations. Rewriting a logarithmic equation as an exponential equation is a useful strategy. Using properties of logarithms is helpful to combine many logarithms into a single one.