IF 2^(3x) = 3 FIND 4^(9x-2)

Answer 1

We can say 4 = 2² and then write the expression as:

Then, because the subtraction of two exponents means the division of the numbers we can say:

The numerator becomes 2^18x and the denominator is 2⁴ = 16.

Using the given equation 2^3x = 3 we can raise both sides to the 6th power giving us:

2^18x = 3⁶ = 729 so now we can substitute 729 into the other equation for 2^18x and get:

4^{9x - 2} = 729/16

Answer 2

2^3x=3

4^(9x-2)=

(2^(2))^(9x-2)=

2^(18x-4)=

2^(18x)*2^(-4)

2^(18x)=(2^(3x))^6

3^6=729

729*2^(-4)=

729/16=

45.5625

You can also get the answer by using logs but I think the teacher wants you to do it this way !

Using logs you get 45.5624999999...=45.5625

Answer 3

Take log of both sides so you can pull the exponent down in front

log 2^(3x) = log 3

3x• log 2 = log 3 divide by 3 • log 2

x = (log 3) / (3 • log 2) approximately .5283298336

substitute in using log

4^[ 9 ( log 3 / (3• log 2)) - 2]

I get 45.5625

Answer 4

Converting this to a log you can solved for x

log₂3=3x

Divide both sides by 3

x=0.5283

Plug this into the second equation

4^9(0.5283)-2 = 45.5625

Answer 5

Use rules of exponents:

2^(9x) = (2^3x)^3 = 3^3 = 27.

4^(9x) = (2*2)^(9x) = 2^(9x) * 2^(9x) = 27^2.

So, 4^(9x-2) = 4^(9x) / 4^2 = 27^2 / 4^2 = (27/4)^2