Re: LOG
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Re: LOG
>Can problems with logs be solved directly?
>Like log7 4 = 2 log7 5 + log7 x+2
>Or almost?
That's a "trick" problem, made up so that if you simply understand
what it is saying the answer "falls out". If the left side of the equation
(one term) is L and the right side (two terms) is R, then it must be true that
7^L = 7^R. The definition of log is the inverse of exponential, so this
becomes just 4 = 5^2 * (x+2).
Solve for x. Bingo!
That's the kind of problem designed to be solved _without_ a calculator, and it
doesn't make sense to ask whether a calculator is capable of solving it. What
is the point of asking such a problem at all? Well, basically just to check
whether you understand the idea about logs and exponentials being inverses.
If you have a calculator in your hand, then your instructor ought to be asking
you to solve equations like log7(x+2) = 3. Or logx 7 = -2 .
Or try this one: log2 x + log3 x = 5 .
The TI-83 would be a good tool to use to solve this one!
RWW Taylor
National Technical Institute for the Deaf
Rochester Institute of Technology
Rochester NY 14623
>>>> The plural of mongoose begins with p. <<<<