# How to Solve Exponential Equations

Exponential equations occur in many types of mathematical modeling problems such as compound interest, cell culture growth, and radioactive decay. There may arise situations in which you need to find the intersection of two exponential equations, that is, solve equations of the form

*ab ^{x} = cd^{x}*

The method and examples below will show you to solve these equations, or you can use the exponential equation solving calculator on the left.

**(Step 1)**First take the logarithm of both sides of the equation. You can use either Log

_{10}or the natural log function Ln. In this instance, we will use Ln. Thus,

Ln(

*ab*) = Ln(

^{x}*cd*).

^{x}**(Step 2)**Use properties of logarithms to break down both sides into simpler expressions.

Ln(

*a*) + Ln(

*b*) = Ln(

^{x}*c*) + Ln(

*d*)

^{x}Ln(

*a*) +

*x*Ln(

*b*) = Ln(

*c*) +

*x*Ln(

*d*)

Now the equation is linear in

*x*.

**(Step 3)**Use algebra to solve for

*x*

*x*Ln(

*b*) -

*x*Ln(

*d*) = Ln(

*c*) - Ln(

*a*)

*x*[Ln(

*b*) - Ln(

*d*)] = Ln(

*c*) - Ln(

*a*)

*x*= [Ln(

*c*) - Ln(

*a*)]/[Ln(

*b*) - Ln(

*d*)]

*x*= Ln(

*c*/

*a*)/Ln(

*b*/

*d*)

There will be a unique solution for

*x*so long as

*a*and

*c*have the same sign, and

*b*and

*d*are both positive and not equal to each other.

**Example 1:**Marty invests $5000 in an account that grows by 3.3% per year and Fawn invests $7600 in an account that grows by 2.1% per year. Neither Marty nor Fawn make any deposits or withdrawals from their accounts. After how many years will Marty have more money in the account than Fawn?

The function that models Marty's account is 5000(1.033)

^{x}and the function that models Fawn's account is 7600(1.021)

^{x}, where

*x*is the number of years. To find when the two accounts are equal, we solve the equation 5000(1.033)

^{x}= 7600(1.021)

^{x}. Using the method above, we get

*x*= Ln(7600/5000)/Ln(1.033/1.02), or

*x*= 85.83 years, or about 35 years and 10 months. After this point, Marty's account will always have more money than Fawn's.

**Example 2:**At time = 0, cell culture A occupies 0.20 cm

^{2}of space in a petri dish, and cell culture B occupies 0.97 cm

^{2}in another petri dish. Cell culture A is growing at a continuous rate of 5.2% per hour, while cell culture B is growing at a continuous rate of 2.9% per hour. When the two cell cultures be the same size? What size will that be?

The function that models the growth of cell culture A is 0.20e

^{0.052x}, and the function that models cell culture B is 0.97e

^{0.029x}, where

*x*is the number of hours and e is the constant of continuous compounding, approximately 2.718281828459045. Setting the two equations equal and solving for

*x*gives us

0.20e

^{0.052x}= 0.97e

^{0.029x}

Ln(0.20) + 0.052

*x*Ln(e) = Ln(0.97) + 0.029

*x*Ln(e)

Ln(0.20) + 0.052

*x*= Ln(0.97) + 0.029

*x*( Ln(e) = 1 )

0.023

*x*= Ln(0.97) - Ln(0.20)

*x*= [Ln(0.97) - Ln(0.20)]/0.023 = 68.65 hours, or 68 hours and 39 minutes.

To find the size of the cultures at

*x*= 68.65 hours, plug this value into the equation for culture A or culture B.

0.97e

^{0.029(68.65)x}= 0.20e

^{0.052(68.65)}= 7.10 cm

^{2}

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