Solver Exponential Equation with Two Exponents

Source code of 'Exponential Equation with Two Exponents'

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About mseely: Math tutor, formerly at St. Philip's College, San Antonio, Texas

==section input

An exponential equation (an equation with the variable in exponent position) with two exponential terms is given:

{{{5^(x-2) = 3^(2x)}}}, for example.

*[input coef1=1]**[input base1=5]^(*[input expcoef1=1]*x+*[input expconst1=-2]) = *[input coef2=1]**[input base2=3]^(*[input expcoef2=2]*x+*[input expconst2=0])

==section solution

Given:

$coef1*$base1^($expcoef1*x+$expconst1) = $coef2*$base2^($expcoef2*x+$expconst2)

The key word here is Exponential Equation. That means that the variable that you are solving for is located in exponent position.

The method for solving Exponential Equations is to apply the log function to both sides of the equation, that is, take the log of both sides of the equation.

Even with two exponential parts, as in this problem, the method is the same. Isolate each exponential part on opposite sides of the equal sign. In this problem, the two exponential parts,  $coef1*$base1^($expcoef1*x+$expconst1) and $coef2*$base2^($expcoef2*x+$expconst2), are already isolated on opposite sides of the equal sign.

So apply the log function to both sides of the equation, or "take the log of both sides."

log($coef1*$base1^($expcoef1*x+$expconst1)) = log($coef2*$base2^($expcoef2*x+$expconst2))

Since we are now dealing with logs, this allows the possibility of the exponents containing your variable to "come down" into coefficient position. This is allowed by the Power Rule of logs, since by the Power Rule of logs, log(x^C) can be rewritten as C*log(x).

First, however, we have to remove the coefficients by using the Product Rule of Logs.

log($coef1) + log($base1^($expcoef1*x+$expconst1)) = log($coef2) + log($base2^($expcoef2*x+$expconst2))

Now we can use the Power Rule of Logs.

log($coef1) + ($expcoef1*x+$expconst1)*log($base1) = log($coef2) + ($expcoef2*x+$expconst2)*log($base2)

Now this is a linear equation in one variable, x, which can be solved by earlier methods. That is, isolate x. First, as always when solving a linear equation, clear parentheses.

log($coef1) + $expcoef1*x*log($base1) + $expconst1*log($base1) = log($coef2) + $expcoef2*x*log($base2) + $expconst2*log($base2)

Since x is on both sides of the equation, use the Addition or Subtraction Property of Equality to get both x terms on the same side.

Now, use the Addition or Subtraction Property of Equality to "move" any "non-x" terms to the other side of the equal sign.

$expcoef1*x*log($base1) - $expcoef2*x*log($base2) = log($coef2) + $expconst2*log($base2) - log($coef1) - $expconst1*log($base1)

Now combine the x terms, that is, combine like terms. In this case, it is easier to think of it as factoring out the common x factor, since the numbers are so ugly, involving logs. Remember that all these logs are just numbers, they are just coefficients of x like the 2 in 2x, for example. Still, we do not want to use our calculator to evaluate log(5) or log(3), for example, because we would have to round our answer--we want to leave them as logs and keep them exact.

x($expcoef1*log($base1) - $expcoef2*log($base2)) = log($coef2) + $expconst2*log($base2) - log($coef1) - $expconst1*log($base1)

Now that all x terms have been combined into one x term, continue to use linear equation solving methods to isolate x. Since something is being multiplied by x, we use the Division Property of Equality to divide it away. Remember, always identify the operation that exists, and then "do the opposite" operation.

x = [log($coef2) + $expconst2*log($base2) - log($coef1) - $expconst1*log($base1)] / [$expcoef1*log($base1) - $expcoef2*log($base2)]

Now, to simplify our answer, we use any of the three log rules that may apply to combine all log expressions into the fewest number of log expressions as possible. First always apply the Power Rule of Logs to move any coefficients of logs back up into exponent position.

Then use the Product Rule of Logs or the Quotient Rule of Logs to combine any sums or differences of logs that you see. (The Quotient Rule of Logs says log(a) - log(b) may be rewritten as log(a/b).)

Simplifying our logs by these rules, we get:
*[assign x=log(10,($coef2*$base2^$expconst2)/($coef1*$base1^$expconst1)) / log(10,$base1^$expcoef1/$base2^$expcoef2)]

x = log[($coef2*$base2^$expconst2)/($coef1*$base1^$expconst1)] / log($base1^$expcoef1/$base2^$expcoef2)

Realize that the following must also be done: any exponents present should be carried out, any negative exponents should be re-expressed using positive exponents, and any fractions should be reduced to lowest terms.  This solver program is not designed to do these final simplification steps.

This expression is the exact answer desired. We avoid putting our logs into a calculator to get a decimal answer which would have to be rounded, and therefore be imprecise.  However, if we do want a decimal approximation, we can use a calculator to compute our logs and get:

x = $x

To summarize, this was an Exponential Equation since our variable was in exponent position, which is always solved by isolating the exponential part of the equation as necessary, and taking the log of both sides. Then one uses strategies for Solving a Linear Equation, not being afraid of ugly expressions like log(5), which is just a number.

==section output
x
==section check
 coef1=1 base1=5 expcoef1=1 expconst1=-2 coef2=1 base2=5 expcoef2=0 expconst2=1 x=3