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

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
equation to solve is:

log(base a)(square root of 16-x^2)

this would be written as:

log(a,(sqrt(16-x^2))) which looks like log%28a%2C%28sqrt%2816-x%5E2%29%29%29

since sqrt(b) is the same as b^(1/2), this equation can be changed to look like:

log%28a%2C%28%2816-x%5E2%29%5E%281%2F2%29%29%29

since log(m^n) = n*log(m), this equation can be changed to look like:

%281%2F2%29+%2A+log%28a%2C%2816-x%5E2%29%29

since 16-x^2 is equal to (4-x)*(4+x), this equation can be changed to look like:

%281%2F2%29+%2A+log%28a%2C%28%284-x%29%2A%284%2Bx%29%29%29

since log(m*n) = log(m) + log(n), this equation can be changed to look like:

%281%2F2%29+%2A+%28log%28a%2C%284-x%29%29+%2B+log%28a%2C%284%2Bx%29%29%29

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:

log%28a%2C%28sqrt%2816-x%5E2%29%29%29 which turns out to be equal to .42254902

final equation is:

%281%2F2%29+%2A+%28log%28a%2C%284-x%29%29+%2B+log%28a%2C%284%2Bx%29%29%29 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 * (... + ...)

you also need to know that root%28n%2Cx%29 is the same as x%5E%281%2Fn%29