allometry equation, y=bx^a help, please.?

Question

how do i turn this equation, y = bx^a into log y = log b + a log x

what properties of log do i use to turn the 1st equation into the 2nd one? thanks for the help.

Answer 1

You are going to use two properties of the log function:
first, use log(m*n) = log(m)+ log(n)
in the first equation, let m= b and n = x^a
then log(bx^a) = log(b) + log(x^a)

next, use log(m^n) = nlogm
I'm going to let you finish this because there is only one more step!

Answer 2

y = bx^a

First, take the log of both sides:

log y = log (bx^a)

Now, remember that log (pq) = log (p) + log (q). If we take p = b and q = x^a in this equation, we get log (bx^a) = log (b) + log(x^a); so the following is true:

log y = log (b) + log (x^a).

Now, remember that log (p^q) = q log p. So, in this case, log(x^a) = a log x. If we make this substitution on the right, we get

log (y) = log (b) + a log (x),

which is what you wanted to show.

Answer 3

ok easy,

y=bx^a

so when you multiply you add on log is crazy so it will be:

log y=log b + log x^a then the a goes to the front is a formula

log y=log b + a log x there it is

Answer 4

y = bxª
by taking the log for each side
log y = log(b* xª)
log y = log b + log xª
log y = log b + a log x

Answer 5

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