Question 174244
Think of the problem as if there were no exponents involved
(a)
{{{x^4 = x*x*x*x}}}
{{{x^5 = x*x*x*x*x}}} so,
{{{x^4*x^5 = x*x*x*x*x*x*x*x*x}}}
This is {{{x}}} multiplied by itself {{{9}}} times, so
{{{x^4*x^5 = x^9}}}
Aha! You just add the exponents
You can do any problem this way & discover the rule
(b)
{{{z^9 = z*z*z*z*z*z*z*z*z}}}
{{{z^10 = z*z*z*z*z*z*z*z*z*z}}}
{{{z^9/z^10 = z*z*z*z*z*z*z*z*z / (z*z*z*z*z*z*z*z*z*z)}}}
There's 1 more {{{z}}} in the denominator, I'm left with
{{{z^9/z^10 = 1/z}}} and I know
{{{1/z = z^-1}}}
Aha!, I just do {{{9 - 10 = -1}}}
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As for logs, just do a very simple experiment
How about {{{log(2,16) + log(2,64)}}}
Big fact: ***** Logs are exponents!!! *****
So I'm adding 2 exponents here
The 1st term says: The exponent to the base {{{2}}} that gives me {{{16}}}
The 2nd term says: The exponent to the base {{{2}}} that gives me {{{64}}}
In the 1st case it's {{{4}}} because {{{2^4 = 16}}}
In the 2nd case it's {{{6}}} because {{{2^6 = 64}}}
So, now I,ve got
{{{log(2,16) + log(2,64) = 10}}}
Remember that {{{10}}} is the sum of 2 logs, so it's a log.
Express it as log(2,?) or "log to the base {{{2}}} that gives me what?
If I raise {{{2^10 = 1024}}}, so
{{{log(2,16) + log(2,64) = log(2,1024)}}}
A better way to write this is:
{{{log(2,2^4) + log(2,2^6) = log(2,2^10)}}}
Aha! Add the exponents
These can be brain-twisters, and it helps if you've
wrestled aligators, but there's no easy way out.