How do you combine #\log _ { 9} 36- \log _ { 3} 2+ \log _ { 0.5} 4#?

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Answer 1 · 2017-08-28T00:06:57

#-1#

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) = v#

Means that #w^v = z#

So what we have is that

#9^a = 36#, #3^b = 2# and #0.5^c = 4#

Take logs of both sides:

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

Now use the exponent rule of logs, ie

#log(w^z) = 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#