If log 9=2, how can i find the value of log 27 ?

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To start with the base 10 log of 9 is not 2 but for the purposes of this question I suspect you mean logs with a base of 3. Regardless we'll solve the problem:

log 9 = 2

log (3^2) = 2

Since log (x^a) = a*log x:

2*log 3 = 2

log 3 = 1 (1)

Now we can convert log 27 into terms of log 3:

log 27 = log (3^3) = 3*log 3

Since we have already established in (1) that log 3 = 1:

3*log 3 = 3*1 = 3

**log 27 = 3**

We'll re-write log 27 using product rule:

loga*b = log a + log b

log 27 = log (3*9) = log 3 + log 9

But log 9 = 2 => log 27 = log 3 + 2 (1)

In the expression of log 27 there is still log 3. We notice that we can write log 3 with respect to log 9 = 2.

We'll write log 9 = log 3^2. We'll use the power rule of logarithms:

log a^b = b*log a

log 9 = 2*log 3

But log 9 = 2 => 2*log 3 = 2

We'll divide by 2 both sides:

log 3 = 1 (2)

We'll susbtitute (2) in (1):

log 27 = 1 + 2

log 27 = 3

The value of log 9=2.

log 9 = log 3^2.

Use the property of logarithm log a^b = b*log a

log 9 = 2*log 3 = 2

log 3 = 1

log 27 = log 3^3

= 3*log 3

= 3*1

= 3

The required value of log 27 = 3

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