How to Solve Logarithmic/Exponential Functions

How to Solve Logarithmic/Exponential Functions - How it Works - Video

Definition of Loga

Definition of loga:

Remember one definition of logarithms is how many of this number do we multiply to get that number using powers?

Let's take a quick look at 2 * 2 * 2 = 8.  We can rewrite 2 * 2 * 2 as 23 because of rules of exponents. Now we have 23 = 8. 23 is our exponential. Now looking back our definition of logarithms. It takes 3 threes multiplied together to get 8 or log2(8) = 3.

Definition of Logarithm and Natural Logarithm

Definition of Logarithm and Natural Logarithm:

With logarithms, the standard base is 10. With natural logarithms, the standard base is e.

Example 1

Example 1:

Here we have log3(2x - 3) = 4. 

Are the bases the same? Do we have multiple terms with the same base? Do we need to use the laws of logarithms?

This time, we have a log that equals a number so we can convert this log function into an exponential function so we can solve for x a little bit easier.

Now that we have our initial input value, 42 . We have to verify that it fits the requirements of any log(x), x > 0. The x in the inequality is the final input value. That might be a bit tricky since the letters are the same. Let's verify to see if the final input value fits the requirement.

Using our initial input value of 42, we found that the final input value of 81. Since 34 = 81 and our final input value is positive, our math is sound. 

Example 2

Example 2:

Here we have log7(x - 2) = log7(4x - 5). 

Are the bases the same? Do we have multiple terms with the same base? Do we need to use the laws of logarithms?

This time, we have two logs that have the same base. So we can set what is inside equal to each other and solve.

Now that we have our the initial input value, 1. We have to verify it that it fits all the requirements of any log, x > 0. The x in the inequality is the final input value, after we simplify what we have in the parentheses. Let's verify to see if the final input value fits the requirement.

Now that we have our initial input value, 1 . We have to verify that it fits the requirements of any log(x), x > 0. The x in the inequality is the final input value. That might be a bit tricky since the letters are the same. Let's verify to see if the final input value fits the requirement.

Using our initial input value of 1, we found that the final input value of -1. Although log7(-1) = log7(-1) is true just looking at it, the final input value cannot be negative, since one definition of log is log(x) for every x >0.

That is why we need to verify each number we find for our initial input value. The initial input value can be negative or positive. It depends the numbers and variables inside the parentheses.

It would make more sense to have our question log7(q - 2) = log7(4q - 5) to avoid confusion with the letters, but that is math for you.

Example 3 Part 1

Example 3 Part 1:

Here we have e2*ln(4x+1) = 9. 

Are the bases the same? Do we have multiple terms with the same base? Do we need to use the laws of logarithms?

This time, we have an exponential equals a number so we can convert this log function into an exponential function so we can solve for x a little bit easier. Luckily we have natural log in our exponent. Since exponentials and logs are inverse, they cancel. Let's solve for x.

Now that we have our initial input values, 1/2 and -1 . We have to verify that both fit the requirements of any log(x), x > 0. The x in the inequality is the final input value. That might be a bit tricky since the letters are the same. Let's verify to see if the final input value fits the requirement.

Using our initial input value of 1/2 and -1, we found that the final input value of 3 and -3. Although both have 9 = 9, only one of them follows the requirements. Which one is it?  

Let's look at why we should stop where the red circle is.

The final input value is -3. Remember the definition of log is log(x) for every x >0. So -1 does not work. Although it looks like it works, properties of logs are a bit different. 2ln(-3) = 1ln(-3) + 1ln(-3) = ln(-3) + ln(-3). We have two inputs that are negative, and the inputs of logs must be greater than 0.

Example 3 Part 2

This is where our work should stop. It can a little tricky, but you should always have the definitions in mind.

Live Worksheet

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