Posted by marcus on Tuesday, November 20, 2012 at 10:48am.
Usually you take logs to the base of the number having exponents. In your example, you have e^(3x-7), so take logs base e.
If you had written 2^(x+3) * 2^(x-7) = 32 then you'd take logs base 2.
As for graphing, just recall the all exponent/log graphs look the same, except for their slopes.
e^x looks just like 10^x. They both go through (0,1). e^x goes through (1,e) and 10^x goes through (1,10).
Fractional bases are just reflected through the y-axis, since (1/2)^x = 2^-x.
Now, logs look just like exponentials, but reflected through the origin, since if
y = e^x,
x = ln y
fractional bases are reflected through the x-axis.
So, as for log_1/2(x) It looks like log_2(x) but upside-down.
Visit wolframalpha.com and enter
log_4 x
or whatever base you want. It will also remind you that log_4(x) = lnx/ln4.
This also explains the reflection using fractional bases. log_1/2(x) = lnx/ln 1/2, and ln 1/2 is negative.
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