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I need answers to these examples
- Thread starter wizkid666
- Start date
I have a so called "long term" sub and he has no idea what the heck he is doing.
1-log5(x-2)=2
8-2e^-x=4
log(x^2+3)=log(x+6)
7^x+3=e^x
(2)(49^x)+(11)(7^x)+5=0
and (4/3)^1-x=5x
Show us what you have done so far so we can see where you are stuck.
I'm pretty much all over the place in pre-calculus at this point...Show us what you have done so far so we can see where you are stuck.
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I get it now. The only thing that was tripping me up was the negative sign in front of the log.In general, few people here will "do" your problems for you. What most will do is to give you a hint on how to start if you say that you are not sure which way to start or will help you if you are stuck if you show your work up to where you are stuck and say something like Now what?
Also read the Read Before Posting.
Now let's take your first problem and work through it. Then you see what you can do on the rest. If you get stuck, come back and show your work. OK?
Log functions and exponential functions are closely related; in fact they are inverses. The thing to remember is that they are functions and can be treated like any other functions. They do have special properties which you use when you must or when it is helpful.
\(\displaystyle 1 - \ logbase5(x-2) = 2\)
Let's isolate the function.
\(\displaystyle logbase5(x - 2) = 1 - 2 = - 1.\)
Now here is the basic property about logarithms; the inverse function is an exponential.
\(\displaystyle logbasea(b) = c \ MEANS \ b = a^c.\)
So
\(\displaystyle x - 2 = 5^{-1} = \dfrac{1}{5}.\)
\(\displaystyle x = 2 + \dfrac{1}{5} = \dfrac{11}{5}.\)
Now let's check our answer.
\(\displaystyle 1 - \ logbase5(\dfrac{11}{5}-2) = 1 - logbase5(\dfrac{1}{5}) = 1 - (-1) = 1 + 1 = 2.\)
Is this clear? If so, try the others remebering that logs and exponentials are inverses.