Exponential equations in which the unknown occurs just once

To solve this type of equation, follow these steps:
• Identify the exponential.

• Invert the operations that were applied to the exponential in the reverse order in which they were applied. The result is that the exponential stands alone on one side of the equation, which now has the form f = a, where the exponent f contains the unknown x.

• If the base of the exponential is e then take natural logarithms of both sides of the equation. Otherwise you may as well take logarithms in base 10. (Do not take logarithms in any other base because your calculator cannot evaluate them.) Immediately use property 3 of logarithms to “bring down” the exponent. This puts the equation into one of these forms:
f · ln (b) = ln (a)   or   f · log (b) = log (a).
• In either case exponents are no longer involved. Finish solving for the unknown, x, by using the basic steps for solving equations.

• Check the solution.

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:
• Make sure that the equation is of precisely the form
a · b x = c · d x.
There can only be two terms and one must be on each side of the equation.

• If the base of either exponential is e then take natural logarithms of both sides of the equation. Otherwise you may as well take logarithms in base 10. (Do not take logarithms in any other base because your calculator cannot evaluate them.) Immediately use property 3 of logarithms to “bring down” the exponent on both sides of the equation. This puts the equation into one of these forms:
ln (a) + x · ln (b) = ln (c) + x · ln (d)   or   log (a) + x · log (b) = log (c) + x · log (d).
• In either case this is now a linear equation in x. It is solved by collecting x terms on the left-hand-side and factoring out x and collecting constant terms on the right:
x (log (b) − log (d)) = log (c) − log (a)
and then isolating x: • Check the solution.

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:
• Identify the exponentials in the equation and make sure that they are all identical. Suppose that they are all equal to f, where the exponent f is an expression containing the unknown, x.

• Use an alias. Replace each occurrence of f with, say Q. This may turn the equation into a linear or fractional or some other non-exponential type of equation in the variable Q. Solve this equation for Q using the usual techniques. Then substitute back f for Q. At this point the equation has the form f = a, where the exponent f contains the unknown x.

• If the base is e then take natural logarithms of both sides of the equation. Otherwise you may as well take logarithms in base 10. (Do not take logarithms in any other base because your calculator cannot evaluate them.) Immediately use property 3 of logarithms to “bring down” the exponent. This puts the equation either into one of these two forms:
f · ln (b) = ln (a)   or   f · log (b) = log (a).
• In either case exponents are no longer involved. Finish solving for the unknown x by using the usual techniques.

• Check the solution.

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:
• Identify the exponentials in the equation and make sure that naming one of them Q makes all the others become positive integer powers of Q. (This may well be the hardest step!)

• Use the alias. Let b x be Q and express all the other exponentials in terms of Q as well. This hopefully turns the equation into a quadratic or polynomial or some other non-exponential type of equation in the variable Q. Solve this equation for Q using the techniques for that type of equation. Then substitute back b x for Q. At this point the equation (or equations) have the form b x = a.

• If the base is e then take natural logarithms of both sides of the equation. Otherwise you may as well take logarithms in base 10. (Do not take logarithms in any other base because your calculator cannot evaluate them.) Immediately use property 3 of logarithms to “bring down” the exponent. This puts the equation either into one of these two forms:
x · ln (b) = ln (a)   or   x · log (b) = log (a).
• In either case exponents are no longer involved. Finish solving for the unknown x by using the usual techniques.

• Check the solution.

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.

 Algebra Coach Exercises

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