How do you combine \log _ { 9} 36- \log _ { 3} 2+ \log _ { 0.5} 4log936log32+log0.54?

1 Answer
Aug 28, 2017

-11

Explanation:

So what the logarithm function is asking is, what is the exponent that the base needs to be raised to to obtain the value, ie

log_w(z) = vlogw(z)=v

Means that w^v = zwv=z

So what we have is that

9^a = 369a=36, 3^b = 23b=2 and 0.5^c = 40.5c=4

Take logs of both sides:

log(9^a) = log(36)log(9a)=log(36), log(3^b) = log(2)log(3b)=log(2) and log(0.5^c) = log(4)log(0.5c)=log(4)

Now use the exponent rule of logs, ie

log(w^z) = zlog(w)log(wz)=zlog(w)

therefore a = log(36)/log(9), b = log(2)/log(3) and c = log(4)/log(0.5)

Now we can simplify some of these. Let's start with a.

a = log(36)/log(9) = log(6^2)/log(3^2) = (2*log(6))/(2*log(3)) = log(6)/log(3)

Since we're going to be using b as well it would be helpful if we could get a into a similar form...luckily we can! Using the addition rule of logs:

log(w) + log(z) = log(wz)

So:

a = log(6)/log(3) = log(2*3)/log(3) = (log(2)+log(3))/log(3) = 1 + log(2)/log(3)

That's us got a and b looking nice so now we turn our attention to c.

c = log(4)/log(0.5) = log(2^2)/log(2^-1) = -2

Therefore

a - b + c = 1 + log(2)/log(3) - log(2)/log(3) - 2 = -1