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.