Solving Exponential Equations
Where x is the independent variable and y is the dependent variable. In many cases, y could also be a constant. e is a constant value of 2.71823 and could actually be any constant value.
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One way of solving this type of equation requires the use of logarithm. You can see the logarithm rules below.
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loga a = 1
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loga 1 = 0
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loga(1/a) = -1
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logam + logan = loga mn
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logam – logan = loga(m/n)
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logaxn = nlogax
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loga(1/y) = -loga y
The base of the logarithm will depend on the value of the number involved. For example, if we have 2x, then we will introduce log2, ;
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…If we have 7x then we will introduce log7
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…and if we have ex, then we will introduce,.. yeah you got it, loge.
The other way of solving exponential equations is by using natural logs i.e; lnx. Ideally, we could introduce a log to any base to solve the question as long as we are consistent. The logic behind introducing a log with similar base as the number involved is so as to make it easier to take logs since the log of a number same as its base is 1.
For example, and so on.
Alternatively, the other way to solve is to simply take natural logs of both sides. The following examples will highlight both techniques.
Natural logs on both sides
The right hand side can be found using a scientific calculator and the answer is approximately 2.9957323.
For the left hand side, earlier, we have shown that the ln(ex) = x. therefore, the above becomes
You can confirm this answer by using your calculator to find e2.9957323 to get 20
Now, what if we had a constant instead of e ? Well the process is similar and you will find that this is easy because the exponential e is actually a constant whose value is 2.71823. Therefore, to solve a problem with a constant instead of e, we follow the same step as above. Let’s explain with an example.
Natural log of 80 is 4.382
Now that the right hand side is solved, how do we solve the left hand side?
We introduce the log rule which states ln(aB) = Blna.
Therefore, the left hand side becomes
Therefore the next step becomes
Now we can find ln 5 using a calculator to give 1.609
We can now solve for x
The introduction of the log law also applies for Example 1. Try it.
Remember we already said that the value of e is ….. yup, you got it! 2.71823. Go on and check from your calculator that the ln of 2.71823 is approximately 1! Therefore, the above becomes
As before!
Harder expressions
In this example, instead of taking the natural log, ln we will take log to base 10 of both sides (be consistent)
Now Log1010 = 1, therefore,
