# If log 9 = 2, how to find the value of log 27

Asked on by schooldump

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

log 9 =2

Let us rewrtie :

9 = 3^2

==> log 3^2 = 2

From logarethim properties we know that:

log a^b = b*log a

==> 2 log 3 = 2

Now let us divide by 2:

==> log 3 = 1

Now we will calculate log 27 :

We will rewrtie :

27 = 3^3

==> log 27 = log 3^3

= 3log 3 =

But we know that log 3 = 1

==> log 27 =   3*1 = 3

Then the answer is : log 27 = 3

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

In this problem we are given that log 9 = 2. Here we will solve the problem without the need for the base of the logarithm arising.

We only use the relation that states: log a^x = x*log a.

Now log 9 = 2

=> log 3^2 = 2

=> 2 log 3 =2

=> log 3 = 2/2

=> log 3 = 1

log 27 = log 3^3

= 3 *log 3

We have already found that log 3 = 1.

Therefore 3 log 3 = 3*1 = 3

Therefore the required value of log 27 is 3.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

log 9 = 2.

To find the value of log27.

Given log 9 = 2.

log3^2 = 2.

2log3 = 2.

log3 = 1.....(1).

log27 = log3^3 , as 27 = 3^3.

log27 = 3log3.

log27 = 3 *1 , as log 3 = 1 from1.

Therefore log(27) = 3, if log 9 = 2.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

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