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# Solving Exponential Equations

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Practice Solving Exponential Equations
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Exponential Equations

When you were first learning equations, you learned the rule that whatever you do to one side of an equation, you must also do to the other side so that the equation stays in balance.  The basic techniques of adding, subtracting, multiplying and dividing both sides of an equation worked to solve almost all equations up until now.  With logarithms, you have more tools to isolate a variable.  Consider the following equation and ask yourself: why is $x=3$ ?  Logically it makes sense that if the bases match, then the exponents must match as well, but how can it be shown?

$1.79898^{2x}=1.79898^6$

#### Watch This

http://www.youtube.com/watch?v=5R5mKpLsfYg James Sousa: Solving Exponential Equations II

#### Guidance

A common technique for solving equations with unknown variables in exponents is to take the log of the desired base of both sides of the equation.  Then, you can use properties of logs to simplify and solve the equation.  See the examples below.

Example A

Solve the following equation for $t$ Note: This type of equation is common in financial mathematics.  This example represents the unknown amount of time it will take you to save $9,000 in a savings account if you save$300 at the end of each year in an account that earns 6% annual compound interest.

$9,000=300 \cdot \frac{(1.06)^t-1}{0.06}$

Solution:

$30 &= \frac{(1.06)^t-1}{0.06}\\1.8 &= (1.06)^t-1\\2.8 &= 1.06^t\\\ln 2.8 &= \ln (1.06^t)=t \cdot \ln (1.06)\\t &= \frac{\ln 2.8}{\ln 1.06} \approx 17.67 \ years$

Example B

Solve the following equation for $x$ : $16^x=25$

Solution:  First take the log of both sides.  Then, use log properties and your calculator to help.

$16^x &= 25\\\log 16^x &= \log 25\\x \log 16 &= \log 25\\x &= \frac{\log 25}{\log 16}\\x &= 1.61$

Example C

Solve the following equation for all possible values of $x$ : $(\log_2 x)^2-\log_2 x^7=-12$ .

Solution:  In calculus it is common to use a small substitution to simplify the problem and then substitute back later.  In this case let $u=\log_2 x$  after the 7 has been brought down and the 12 brought over.

$(\log_2 x)^2-7 \log_2 x+12 &= 0\\u^2-7u+12 &= 0\\(u-3)(u-4) &= 0\\u &= 3, 4$

Now substitute back and solve for  $x$ in each case.

$\log_2 x &= 3 \leftrightarrow 2^3=x=8\\\log_2 x &= 4 \leftrightarrow 2^4=x=16$

Concept Problem Revisited

In the equation, logs can be used to reduce the equation to $2x=6$

$1.79898^{2x}=1.79898^6$

Take the log of both sides and use the property of exponentiation of logs to bring the exponent out front.

$\log 1.79898^{2x} &= \log 1.79898^6\\2x \cdot \log 1.79898&=6 \cdot \log 1.79898\\2x &= 6\\x &= 3$

#### Vocabulary

Taking the log of both sides is an expression that refers to the action of writing log in front of the entire right hand side of an equation and the entire left hand side of the equation.  As long as neither side is negative or equal to zero it maintains the equality of the two sides of the equation.

#### Guided Practice

1. Solve the following equation for all possible values of $x$ : $(x+1)^{x-4}-1=0$

2. Light intensity as it travels at specific depths of water in a swimming pool can be described by the relationship between $i$  for intensity and $d$  for depth in feet.  What is the intensity of light at 10 feet?

$\log \left(\frac{i}{12}\right)=-0.0145 \cdot d$

3. Solve the following equation for all possible values of $x$ .

$\frac{e^x-e^{-x}}{3}=14$

1.  $(x+1)^{x-4}-1=0$

$(x+1)^{x-4}=1$

Case 1 is that $x+1$  is positive in which case you can take the log of both sides.

$& (x-4) \cdot \log (x+1)=0\\& x=4,0$

Note that $\log 1=0$

Case 2 is that $(x+1)$  is negative 1 and raised to an even power.  This happens when $x=-2$ .

The reason why this exercise is included is because you should not fall into the habit of assuming that you can take the log of both sides of an equation.  It is only valid when the argument is strictly positive.

2.  Given $d=10$ , solve for $i$  measured in lumens.

$\log \left(\frac{i}{12}\right) &= -0.0145 \cdot d\\\log \left(\frac{i}{12}\right) &= -0.0145 \cdot 10\\\log \left(\frac{i}{12}\right) &= -0.145\\\left(\frac{i}{12}\right) &= e^{-0.145}\\i &= 12 \cdot e^{-0.145} \approx 10.380$

3. First solve for $e^x$ ,

$\frac{e^x-e^{-x}}{3} &= 14\\e^x-e^{-x} &= 42\\e^{2x}-1 &= 42e^x\\(e^x)^2-42e^x-1 &= 0$

Let $u=e^x$ .

$& u^2-42u-1 = 0\\& u = \frac{-(-42) \pm \sqrt{(-42)^2-4 \cdot 1 \cdot (-1)}}{2 \cdot 1}=\frac{42 \pm \sqrt{1768}}{2} \approx 42.023796, -0.0237960$

Note that the negative result is extraneous so you only proceed in solving for  $x$ for one result.

$e^x & \approx 42.023796\\x & \approx \ln 42.023796 \approx 3.738$

#### Practice

Solve each equation for $x$ .  Round each answer to three decimal places.

1.  $4^x=6$

2.  $5^x=2$

3.  $12^{4x}=1020$

4.  $7^{3x}=2400$

5.  $2^{x+1}-5=22$

6.  $5x+12^x=5x+7$

7.  $2^{x+1}=2^{2x+3}$

8.  $3^{x+3}=9^{x+1}$

9. $2^{x+4}=5^x$

10.  $13 \cdot 8^{0.2 x}=546$

11.  $b^x=c+a$

12. $32^x=0.94-.12$

Solve each log equation by using log properties and rewriting as an exponential equation.

13.  $\log_3 x+\log_3 5=2$

14.  $2 \log x=\log 8+\log 5-\log 10$

15. $\log_9 x=\frac{3}{2}$