Let u = log 2 and v = log 3. Evaluate the following expressions in terms of u and/or v. For example, log 9 = log(32) = 2 log 3 = 2v.

Question

(a) log 6

(b) log 0.08

(c) log √(3/2)

(d) log 5

Answer 1

Here we need to simply use our logarithm identities as per the example.

a) log 6 = log 2•3 = log 2 + log 3 = u + v

b) log 0.08 = log 8/100 = log 8 - log 100 = log 2³ - log 10¹⁰ = 3 log 2 - 10 log 10 = 3u - 10

(NOTE: log (with no base) is log₁₀ so log 10 = log₁₀10 = 1)

c) log √(3/2) = log √3 - log √2 = log 3^1/2 - log 2^1/2 etc

d) log 5 (this one is tricky. We need to get it into terms of 2 and 3 but log (3 + 2) has no identity to separate it so we have to be creative.

log 5 = log 30/6 = log 30 - log 6 = log 3•10 - log 3•2 = log 3 + log 10 - (log 3 + log 2) = log 3 + 1 - log 3 - log 2

= 1 - log 2 = 1 - u

Answer 2

First, let's review an important property of logarithms:

log(x^a) = a·log(x)

In other words, if we are taking the logarithm of a quantity raised to an exponent, then we can pull the exponent out and multiply it by the logarithm.

Ok, so let's return to the Question (c)

log[√3/2]

Well, we have that we are taking the square root of 3/2, and taking the square root of something is the same as raising it to the 1/2 power. So our exponent ("a" from the example above) would be 1/2.

Thus,

log[√3/2] = 1/2 · log[ 3/2 ]

Ok, we are in the home stretch now!!! Another important property of logarithms is that:

log(a/b) = log(a) - log(b)

So, we must have that:

1/2 · log[ 3/2 ] = 1/2 · ( log(3) - log(2) )

Finally, we can substitute in u = log(2) and v = log(3) to get

log[√3/2] = 1/2 · (v - u)