Solving Exponential Functions

Published: Dec 26th, 2020

In the last lesson, we got acquainted with exponential functions. This time around, let's solve some problems!

Solve \(7^{2x-3}=49\) for \(x\)

Let's start by expressing the 49 as an exponent. As you might know, \(7 \times 7 = 49\), so:


Alright, now let's solve for \(x\) in the exponents: 

\(2x-3 = 2\)

\(2x = 5\)

\(x = \frac{5}{2}\)

That wasn't so bad, right? Let's spice it up.

Solve \(5^{x^2-4x}=\frac{1}{125}\) for \(x\) 

 As you might remember, we can interchange between an inverse and a negative exponent. In this case, we can rewrite \(\frac{1}{125}\) as \(\frac{1}{5^{3}}\), and then as \(5^{-3}\)


As we did before, let's solve for \(x\) in the exponents: 

\(x^2 - 4x = -3\)

\(x^2 - 4x + 3 = 0\)


Therefore, there are two solutions: \(x=3\) and \(x=1\). Onwards!

Solve \((3^{2x})-2(3^{x})+1=0\) for \(x\) 

Wooooo.... this one seems hard, no?

Well, let's look at it closer. As you might remember from the Laws of Exponents, we can rewrite the product of two exponents with the same base as the sum of their powers.

In this case, we can also say that \(3^{x} \times 3^{x} = 3^{2x}\). Interesting.

Now, let's try to simplify this expression by getting rid of \(3^x\). Let's call it \(u\)

\(u^2 -2u+1=0\)

Now this is much more familiar! However, it's VERY IMPORTANT that we remember to re-substitute, because these roots will not represent the true answer!

Factoring this expression, we get:


Therefore, there is only one solution \(u=1\). However, we're not done yet... remember to re-substitute:

\(u = 1\)

\(3^x = 1\)

As you might remember, any number except for zero raised to the power of 0 is equal to one. Therefore,

\(x = 0\)

This one's short, so be sure to try the practice problems to make it really click! 

Now, you might be wondering "what if not all the terms have the same base?" Well, that's the subject of our next lesson: Logarithms!


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