Question 244996
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(a,(sqrt(16-x^2)))}}}


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


{{{log(a,((16-x^2)^(1/2)))}}}


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


{{{(1/2) * log(a,(16-x^2))}}}


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


{{{(1/2) * log(a,((4-x)*(4+x)))}}}


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


{{{(1/2) * (log(a,(4-x)) + log(a,(4+x)))}}}


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(a,(sqrt(16-x^2)))}}} which turns out to be equal to .42254902


final equation is:


{{{(1/2) * (log(a,(4-x)) + log(a,(4+x)))}}} 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(n,x)}}} is the same as {{{x^(1/n)}}}