# Solve for the unknown? 243^(x-3)=9^x

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We have to find x given 243^(x-3) = 9^x

Here we use logarithms to accomplish the same.

243^(x-3) = 9^x

Take the log to the base 3 of both the sides

log (3) [ 243^(x-3)] = log (3) [9^x]

Now log a^b = b* log a and log (a) a = 1.

=> (x - 3) log (3) 243 = x log (3) 9

=> (x - 3) log (3) (3^5) = x log (3) (3^2)

=> (x - 3) * 5 * log (3) 3 = x * 2 * log (3) 3

=> 5*(x - 3) = 2x

=> 5x - 15 = 2x

=> 3x = 15

=> x = 15/3

=> x = 5

**Therefore x is equalÂ to 5**.

We notice that the bases of exponentials from both sides could be written as powers of 3.

243 = 3^5

9 = 3^2

We'll re-write the equation having common bases both sides:

3^5(x-3) = 3^2x

Since the bases are matching, we'll use the one to one property of exponentials and we'll get:

5(x-3) = 2x

We'll remove the brackets from the left side:

5x - 15 = 2x

We'll subtract 2x and add 15 both sides:

3x = 15

We'll divide by 3:

**x = 5**

**The solution of the equation is x = 5.**