# Solve the equation log 2 ( 3+ log 3 x) = 2

hala718 | Certified Educator

log 2 (3+log 3x) = 2

==> 3+ log 3x = 2^2

==> 3+ log 3 x = 4

==> log 3 x = 1

==> x = 3^1 = 3

==> x= 3

Now let us check:

log 2 (3+ log 3 (x) = 2

loh 2 (3 + log 3 (3) = 2

We know that log 3 3 = 1

==> log 2 (3+1) = 2

==> LOG 2 4 = 2

==> 2 = 2

giorgiana1976 | Student

Before solving the equation, we'll impose constraints of existence of logarithms:

3+ log 3 x > 0 => log 3 x>-3

and

x>0

log 3 x>-3

log 3 x>-3*1

log 3 x>-3*log 3 3

log 3 x> log 3 3^-3

log 3 x> log 3 (1/27)

Because the base of logarithms is 3>1, the the function is increasing, so:

x > 1/27

The admissible solutions of x have to belong to the interval (1/27, inf.).

We'll solve the equation:

log 2 ( 3+ log 3 x) = 2

3+ log 3 x = 2^2

3+ log 3 x = 4

log 3 x = 4-3

log 3 x = 1

x = 3^1

x = 3

Because 3 is in the interval (1/27 , inf.) is a valid solution.

tonys538 | Student

The solution of `log_2 ( 3+ log_3 x) = 2` has to be determined.

If `log_b x = y` , `x = b^y` .

Applying this to `log_2 ( 3+ log_3 x) = 2` gives:

`3+ log_3 x = 2^2`

`3+ log_3 x = 4`

`log_3 x = 4 -3`

`log_3 x = 1`

Again applying the relation given earlier, we get

x = 3^1

x = 3

The solution of the equation `log_2 ( 3+ log_3 x) = 2` is x = 3

uniquemiss | Student
can some one help Solve 1) 2r+1P4 if 9Pr=504 2) nCr if r!=5040 and nPr= 604800 3) x+2P3=42x
neela | Student

log2 (3+log3x) =2.

To solve for x.

Solution:

log2 (3+ log3) = 2. Taking antologarithms with respect to base 2, we get:

(3+log3x ) = 2^2 = 4.

log3 x = 4-3 = 1. Take antilog (hope this is 10 base not 3 base).

3x = 10

x = 10/3.

If log3  x  is a log to base e, then

log3 x = 1. Antilog wrt base 3 gives:

x = 3.