log/ exponents

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xc630

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Sep 1, 2005
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The question instructs to use a calcualtor to solve the following problems. However I do not know how to do so on a calculator.

a) log(5) 27

b) 3^x-2 = 30

c) (x-2)^3 = 30

Thanks for your help.
 
a) Use the change-of-base formula you've memorized, convert to common or natural logs, and punch the buttons.

b) Is the power "x - 2", or just the "x"? When you reply, please show how far you've gotten (on paper) in solving the equation for "x=" (which is probably where you'll be using the calculator).

c) Take the cube root of both sides, and add "2" to both sides. To get a decimal approximation, punch the buttons.

Eliz.
 
The power is "x-2". However I don't know how to change the 30 to a base of 3, which i think I need to do.
 
What kind of calculator do you have? If it is intended for this course it should have a log button.
The conversion Eliz is refering to is
log<sub>5</sub> (27) = log(27)/log(5)
 
I have a ti-83 but i got hte 1st and 3rd prblems. I need help with the 2nd problem
 
Take logs of both sides, divide off the "log(3)", and add "2" to both sides.

Eliz.
 
She meant divide both sides by log(3) after you take the log of both sides.
 
log(3^(x-2))=(x-2)log(3)
log(30)=log(30)
or if you want to do it a little differently
log(30)=log(3)+log(10)=log(3)+1
 
Sorry I'm still confused. I have (x-2) log 3= 30.

How did u get log 30 = log 30? What happended to the x-2?
 
I don't know how to say it in print. I was trying to give you each side in a separate equation, then combine them.
Maybe?
The log of 30 is log(30)
The log of 3^(x-2) is (x-2)*log(3)
When you take logs of both sides of
3^(x-2)=30
you get
(x-2)log(3)=log(30)
So you can solve it by Eliz's method to get
(x-2)=log(30)/log(3)
x=log(30)/log(3) + 2
 
Pheeew!
Now that you have that, my alternative answer is
(x-2)log(3)=log(3)+1
x-2=1+1/log(3)
x=3+1/log(3)
Its the same "number" but I have one less log.
------------------
Gene
 
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