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How to solve an Exponential equation?

So, what exactly do you understand by an Exponential equation in Algebra?
When a constant i.e. an integer is raised to the power of something denoted by a variable and is equated to a constant that can be termed as an exponential equation. This is a ‘univariate’ algebraic equation. And very importantly should not be confused with exponential function. A very basic example of this type of equation is given below:

8ⁿ = 64 ; which can be written as,  8ⁿ = 8² ; hence n = 2

Now let’s check the different methods of solving such exponential equations.
There are primarily two methods of solving exponential equations unless you are smartly speeding it up with a calculator. One method is solving it by the definition and the other is solving by logarithm.

Solving by definition:

4 x² – 4x = 256   =>  4x² – 4x = 44   =>   x² – 4x = 4  =>   x² – 4x – 4 = 0    =>   (x – 4)(x + 1) = 0   =>  x = 4, -1

This is one way of solving an exponential equation when you make both sides of the equation comparable to each other which is only possible when you have an identical base i.e. the same constant number as bases on both sides which then implies that the exponents must be equal to each other to hold the equation true.

Now what if the equation is not so simple, which is mostly the case, always?  We need another method to solve those equations when both sides of the equals to sign can’t be brought to comparable figures, ex:   2X = 30, here 30 can’t be converted to a power of 2. Hence here we need a different approach towards the problem. This is when the second method comes in.

Solving by logarithm:

The principle followed here is that as long as we do the same operation on both sides of the equation the result remains same. The variable whose value we need to derive is an exponent here. And technically logarithms are ‘inverse’ of exponents. So let’s take an example here.

6X = 40    i.e.   log(6X) = log(40) 

Now, the log rule says anything that is an exponent can be brought in front as a multiplier. Also, we can use log to the base-10, which is ln , or we can also use log to any other base for example base-6 here, the ultimate result will be same. We will show you how.

So,   x log10(6) = log10(40)  ;   hence  x = ln(40)/ ln(6)

Now, if we change the base of the log:

X log6(6) = log6(40)    ;     x (1) = log6(40) [since logn(n) = 1]  ;   x = log6(40)

Now, to convert log base-6(40) to log base-10(40) we have to divide both of them as shown below:

log6(40) = log10(40)/ log10(6) = ln(40) / ln(6) = x   Hence, as you can see the final result is same as above.