Exponential equations in which the unknown occurs just once

To solve this type of equation, follow these steps:


Example: Solve the exponential equation 2 · 5 x + 3 = 21 for x.

Solution: This equation contains a single exponential, 5 x. (Notice that the number 2 is not part of it!) To isolate the exponential, subtract 3 and then divide by 2 on both sides of the equation. The result is that the exponential is isolated on the left-hand-side of the equation:
5 x = 9.
Take logarithms to base 10 of both sides:
log(5 x) = log(9)
Use property 3 of logarithms to bring down the exponent:
x · log(5) = log(9)
Solve for x:

Exponential equations of the form a · b x = c · d x

To solve this type of equation, follow these steps:


Example: Solve the exponential equation 5 · e 1.7 x = 2 · 4 2.9 x for x.

Solution: Because one of the exponentials has base e, take natural logarithms of both sides of the equation:
ln (5 · e 1.7 x ) = ln (2 · 4 2.9 x ).
On both sides of the equation, use property 1 of logarithms to split up the logarithm of the product
ln (5) + ln (e 1.7 x ) = ln (2) + ln (4 2.9 x ).

On the right-hand-side, use property 3 of logarithms to bring down the exponent. On the left-hand-side you could do the same, but instead you can just use the fact that the natural log and the exponential function cancel. This gives:
ln (5) + 1.7 x = ln (2) + 2.9 x ln (4)
Simplify:
1.609 + 1.7 x = 0.6931 + 4.02 x
This is a linear equation in x. Solve it by collecting x terms on the left-hand-side and constant terms on the right, and then isolating x. The solution is x = 0.3949. Check this solution by substituting it into the original equation. This gives 9.784 = 9.784 so it checks out.



Exponential equations in which the same exponential occurs several times

To solve this type of equation, follow these steps:


Example: Solve this exponential equation for t:
Solution: This equation contains two identical exponentials, ek t. We replace both occurrences of ek t with Q:
This is now a fractional equation in Q. In order to clear the denominators, multiply both sides by 1 − Q:
a (1 − Q) = b (1 + Q).
This is now a linear equation in Q. In order to collect like terms, distribute on both sides:
aa Q = b + b Q.
Collect the constant terms on the left-hand-side. Then collect the terms containing Q on the right-hand-side and factor out Q:
ab = Q (a + b).
Solve for Q:
Now substitute back ek t for Q. The result is a single exponential that is isolated:
We are finally at the point where we can take natural logarithms. Doing so gives:
The unknown, t is no longer in the exponent. Now divide through by −k to solve to t:
We could leave the answer in this form or we could get fancy and use the fact that to get rid of a − sign and write the answer as:

Exponential equations containing exponentials b x, b 2 x, b 3 x, …

Suppose that an exponential equation contains the exponentials b x, b 2 x, b 3 x, etc., where b could be any base. This type of equation can be solved by a slight extension of the alias method used above. It depends on the following observation:
If b x is replaced by the alias, Q, then
b 2 x = (b x ) 2 = Q 2,
b 3 x = (b x ) 3 = Q 3,
b 4 x = (b x ) 4 = Q 4,   etc.
Thus all the exponentials can be replaced with positive integer powers of Q. Even exponentials like b 2 x + 1 can be expressed in terms of an integer power of Q since
b 2 x + 1 = b 1 · b 2 x = b Q 2
Here are the steps to solve this type of equation:


Example: Solve the exponential equation e − 2 t + e − t + 1 − 1 = 0.

Solution: This is a quadratic equation in e − t as can be easily seen if we use the properties of exponents to rewrite it as:
(e − t ) 2 + e 1 · e − t − 1 = 0.
Now replace e − t with the alias Q. The result is:
Q 2 + e Q − 1 = 0.
(Recall that e is the number approximately equal to 2.718.) Solve this quadratic equation for x using the quadratic formula with a = 1, b = e and c = −1:
Now substitute back e − t for Q:
Since we can’t take the logarithm of a negative number the second solution is extraneous. That leaves the only solution:

t = −ln (0.3282) = 1.11



Note that not all exponential equations can be solved using algebra. For example consider the seemingly simple equation x = 10x. We cannot get the x out of the exponent without putting the other x into a logarithm. This equation can only be solved approximately using a computer.


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