You can put this solution on YOUR website!
Any problem where the variable is in an exponent can be solved using logarithms. But if the equation can be rewritten so that each side is a power of the same number, then there is a much easier way. So much easier that it is definitely worth the effort to see if it is possible.
So we will start by seeing if we can take the easy way. The left side is already a power of 3. Can we rewrite the right side so that it is a power of 3, too? Or can we rewrite both sides so that they are powers of some other number? If you know or figure out that and if you understand negative exponents then you will find that . So we can use the easy path:
The only way two powers of 3 can be equal (as this equation now says they are), is if the exponents themselves are equal, too. So:
Now we solve this. Add 3:
Factoring:
(x+1)(x-1) = 0
Zero Product Property:
x+1 = 0 or x-1 = 0
Solving:
x = -1 or x = 1
P.S. We could have solved using square roots:
but this way we have to remember the positive and negative square roots. This is easily forgotten. By solving the way I did above, we don't have to remember. The procedure automatically handles the two square roots.
P.P.S. Not only is using logarithms harder than the way we did this problem but it often results in answers that are slightly off. When solving with logarithms you often use your calculator to find some logarithms. Most logarithms you get from a calculator are (very close) decimal approximations of the right numbers. So if we had used logarithms to solve this problem we might have ended up with 1.000000001 or -0.9999999998 instead of the exactly correct 1 and -1. The moral of this story is: Solve these problems the way we did this one whenever you can. It is easier and more exact. But some problems, like , cannot be done this way. Use logarithms on these problems.
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