### Assignments:

Unfinished Assignment Study Questions for Lesson 28

### Lesson Objectives:

- Base-Exponent Property
- Property of Logarithmic Equality
- Solve Exponential Equations
- Solve Logarithmic Equations

Base-Exponent Property.

For any b > 0, where b != 1, b^x = b^y, if x = y.

Solve the exponential equation.

5^(2x-7) = 125. 125 is the same as 5^3, so we can write each side as a power of the same number. 5^(2x-7) = 5^3. So using the Base-Exponent Property, we can set the exponents equal to each other. 2x-7 = 3. Add 7 to both sides, 2x = 10. Divide both sides by 2, and we get x = 5.

Now let's check if x = 5 is the solution. 5^(2(5)-7) = 125. 5^(10-7) = 125, or 5^3 = 125. So we get 125 = 125, a true statement. So, x = 5 is our solution.

Property of Logarithmic Equality.

For any value M > 0, and N > 0, and b > 0, where b != 1, log_bM = log_bN, if M = N.

Solve the exponential equation.

85^x = 70. Start by taking the common log of both sides. So we have log85^x = log70. This can be simplified to x * log85 = log70. And then if we divide both sides by log85, we get x = (log70)/(log85). This is equal to approximately 0.956.

Solve the exponential equation.

e^x+4e^(-x) = 5.

Start by rewriting everything with positive exponents. e^x+4/(e^x) = 5.

Now multiply both sides by e^x. So we get e^(2x)+4 = 5e^x.

Now let's move all of our terms over to one side. So we have e^2x-5e^x+4 = 0. If we let u = e^x, then we could rewrite this as u^2-5u+4 = 0.

We can factor this: u*u gives us u^2, and -4*-1=4, but adds to -5. So u = 4 or positive 1.

Now we have to substitue e^x back in for u. So we have e^x = 4 and e^x = 1.

We have to take the natural log of both sides, so we have ln(e^x) = ln4. That means that x = ln(4). And if we do ln e^x = ln1, we have x = ln1.

So the solution is x = ln4 or 0, or x is about 1.386 or 0.

Solve the logarithmic equation.

log(x)+log(x+15) = 2. So let's start by using the product rule to obtain a single logarithm. We get log(x(x+15)) = 2, or log(x^2+15x) = 2. Now, since we don't have any base written, it's understood that this is a common log, or a log_10.

So we can write the equivalent exponential equation as 10^2 = x^2+15x, or 100 = x^2+15x. Now let's put everything over to one side. So we get 0 = x^2+15x-100. And we can factor this and get zero is equal to: x*x gives us x^2, and positive 20, and -5 multiply to make -100, but add to make positive 15.

So we get x = -20 or positve 5. And we'll need to check our answer.

x = -20. So we'll plug that in for x. log(-20)+log(-20+15) = 2. Now we know that -20 is not a solution, because negative numbers do not have real number logarithms.

Now let's check x = 5. log(5)+log(5+15) = 2, or log(5)+log(20) = 2. This is the same as log(5*20) = 2, or log(100) = 2. So we can rewrite this as log_(10)10^2 = 2. So we get 2 = 2, a true statement. That means that x = 5 is the solution.