SOLUTION: solve the equation exactly for t. 64e^0.18t = 25(1.4)t Help!! thank yoouuu

Question 350030: solve the equation exactly for t.
64e^0.18t = 25(1.4)t
Help!! thank yoouuu
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
I assume the problem is

If not, then repost your question.

To solve an equation where the variable is in one or more exponents, we usually use logarithms. We will start by finding the logarithm of each side. But logarithm of which base? Answer: It doesn't really matter! But

the final, simplified exact answer will be simpler if we choose a base of logarithm that matches one of the bases with an exponent. In this problem that means we will get a simpler expression if we choose base e (aka ln) or base 1.4 logarithms.
if you want (or may want) a decimal approximation, you may want to choose a base of logarithm your calculator "knows". Most calculators "know" base 10 and/or base e logarithms.

Since base e logarithms will make the exact solution simpler and give us the option of finding a decimal approximation, if desired, we will use base e.

Next we will use properties of logarithms to manipulate these expressions. First we will use the property to separate the coefficients into separate logs:

Next we can use the property to move the exponent out in front. (This is the key to solving the problem this property allows us to move the exponents, where the variable is, out in front (not as an exponent) where we can then use regular Algebra to solve for the variable.)

By definition, ln(e) = 1. (This is why choosing base e makes this solution simpler. We end up with one less logarithm.) So the equation simplifies to:

Now that the variable is out of the exponent, we can solve for it. To do so we start by getting the variable terms on one side and the other terms on the other side of the equation:

Now we factor out the t on the left side:

And divide both sides by (0.18 - ln(1.4)):

This is an exact solution to your equation and it may be considered to be in simplified form. The only other thing you could do is use the property of logarithms to combine the logarithms in the numerator:

(I honestly don't know which equation is "simpler".)

If you need a decimal approximation, then you just find the three logarithms with your calculator and simplify.

Just so you can see how a different choice of base for the logarithms works out, I'll pick a base at random, let's say 12, and solve it with this base. The steps and logic is all the same as above (except we won't have a logarithm "disappear" like ln(e)) so the solution will be done without comment.

This is also an exact solution to your equation. As you can see, it is not quite as simple as our earlier solution. And to find a decimal approximation for the solution using the base 12 logs, we would have to use the base conversion formula, to convert all the base 12 logs into base 10 or base e logarithms.
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