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RE: Exponents: getting the value of x.
Posted:
Sep 17, 2007 1:48 PM
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All,
I notice there is a correction to the last line in the solution of 2^x =
16 in my previous e-mail. I wrote "x = log 2 / log 16". It should have
read, or course:
x = log 16 / log 2,
which is what yields 4 if you put it into your calculator. Sorry for the
error.
Cindy
-----Original Message-----
From: numeracy-approval@world.std.com
[mailto:numeracy-approval@world.std.com] On Behalf Of Martha Haehl
Sent: Wednesday, September 12, 2007 11:08 AM
To: numeracy@europe.std.com
Subject: RE: Exponents: getting the value of x.
The second question requires a calculator only if you are getting a
decimal approximation. The exact answer to is X = log 15/ log 2.
Martha
Martha Haehl,
Ph.D., Education & Urban Leadership
Mathematics Instructor
816-759-4221
martha.haehl@mcckc.edu
Metropolitan Community College-Penn Valley
3201 S.W. Trafficway
Kansas City, MO 64111
>>> "Ravitsky, Cindy" <cravitsk@cotc.tec.oh.us> 9/11/2007 1:52:41 pm >>>
For exponential equations such as these, you can solve for x by taking
the log (common or natural) of both sides, then using a property of
logarithms to get the x out of the exponent. For example:
2^x = 16
log 2^x = log 16
x log 2 = log 16
x = log 2 / log 16
If you punch this into a calculator, it gives you 4.
This same technique can be used on 2^x = 15 to get
X = log 15/ log 2
X = 3.9069 (approximately)
This of course, requires use of a calculator or log tables.
Cindy
-----Original Message-----
From: numeracy-approval@world.std.com
[mailto:numeracy-approval@world.std.com] On Behalf Of Ed Wall
Sent: Saturday, September 08, 2007 6:48 PM
To: numeracy@europe.std.com
Subject: Re: Exponents: getting the value of x.
An interesting question. A comparable question, in a sense, is how do
you get x in 2 times x = 8 without using a calculator. A, perhaps,
more interesting question is how do you get x in
2^x = 15
without a calculator?
Ed Wall
>2^4 = 16
>
>now
>
>2^x = 16
>
>we all know that x = 4. But how exactly do we get x without using a
>calculator?
>
>
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