SOLUTION: write the expression as a sum or difference of logarithms.
log(base a)(square root of 16-x^2)
^
so far ive tried log(base a) 16-x^2)^1/2...and then 1/2 log(base a)(16-x^2)...but
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-> SOLUTION: write the expression as a sum or difference of logarithms.
log(base a)(square root of 16-x^2)
^
so far ive tried log(base a) 16-x^2)^1/2...and then 1/2 log(base a)(16-x^2)...but
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Question 244996: write the expression as a sum or difference of logarithms.
log(base a)(square root of 16-x^2)
^
so far ive tried log(base a) 16-x^2)^1/2...and then 1/2 log(base a)(16-x^2)...but i dont thinki can do this 1/2log(base a)16-1/2log(base a)x^2?? Answer by Theo(13342) (Show Source):
since sqrt(b) is the same as b^(1/2), this equation can be changed to look like:
since log(m^n) = n*log(m), this equation can be changed to look like:
since 16-x^2 is equal to (4-x)*(4+x), this equation can be changed to look like:
since log(m*n) = log(m) + log(n), this equation can be changed to look like:
I don't believe it can be transformed any further.
to confirm this answer is good, then substitute some values for a and x and see if the original equation and the final equation give you the same answer.
let a = 10 (calculator can do logs to the base 10)
let x = 3 (random selection of not too big a number and small enough so the square root doesn't go negative)
original equation is:
which turns out to be equal to .42254902
final equation is:
which turns out to be .42254902
these answers are the same so it looks like the transformed equation is accurate.
there are 3 basic transformation formulas you need to know.
log(m^n) = n*log(m)
log(m*n) = log(m) + log(n)
log(m/n) = log(m) - log(n)
you also need to make sure that the multiplier works on all that needs to be multiplied.
in our case we had the equivalent of:
log((a*b)^c)
this became c * log(a*b)
then this became c * ( log(a) + log(b))
the c had to multiply them both as I did above with the 1/2 * (... + ...)