Friday, April 2, 2010

reversing log

I was tutoring, and one particular type of problem came up as difficult, solving for a variable in a log.

2 = \log x
To solve this, first exponent both sides on 10, which will reverse the log. The hard part seems to be here.
10^2 = 10^{\log x}
and simplify (10^{\log x} = x).
100 = x

To make is harder, multiply the x by a constant inside the log.
2 = \log 2x
put both sides as exponents of 10
10^2 = 10^{\log 2x}
100 = 2x
divide both sides by 2
\frac{100}{2} = \frac{2x}{2}
50 = x

and this time add something to the x inside the log
2 = \log\left( 2x + 1 \right)
put both as exponents of 10
10^2 = 10^{\log\left( 2x + 1 \right)}
100 = 2x + 1
subtract 1 from both sides
100 - 1 = 2x + 1 - 1
99 = 2x
divide both sides by two
\frac{99}{2} = \frac{2x}{2}
\frac{99}{2} = x

what can I do to make it harder? Two layers of log? two layers of logs of different bases? that should be plenty hard.

2 = \ln\log x
The \ln x is the outside log, so it needs to be dealt with first, make both sides exponents of e
e^2 = e^{\ln\log x}
e^2 = \log x
make both sides exponents of 10
10^{e^2} = 10^{\log x}
10^{e^2} = x

And now for one that looks a bit scary, the log is in an exponent.
2 = e^{\log x}
First there is the e^x to deal with, so take the natural logarithm of both sides.
\ln 2 = \ln e^{\log x}
\ln 2 = \log x
and put both sides on an exponent of 10, as in the previous solutions
10^{\ln 2} = 10^{\log x}
10^{\ln 2} = x

It is all a matter of removing one layer at a time to get to the thing being solved for.

all good?

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