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}

simplify

100 = 2x

divide both sides by 2

\frac{100}{2} = \frac{2x}{2}

simplify

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)}

simplify

100 = 2x + 1

subtract 1 from both sides

100 - 1 = 2x + 1 - 1

simplify

99 = 2x

divide both sides by two

\frac{99}{2} = \frac{2x}{2}

simplify

\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}

simplify

e^2 = \log x

make both sides exponents of 10

10^{e^2} = 10^{\log x}

simplify

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}

simplify

\ln 2 = \log x

and put both sides on an exponent of 10, as in the previous solutions

10^{\ln 2} = 10^{\log x}

simplify

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?