If $3^{33}+3^{33}+3^{33}=3^{x}$. Solve for $x$.

Question

If $3^{33}+3^{33}+3^{33}=3^{x}$. Solve for $x$.

So we have:

$$3^{33}+3^{33}+3^{33}=3^{x}$$

I added the left side and obtained:

$3(3^{33})=3^{x}$

The problem I have is that extra $3$. If not, I could have said $x=33$. Any hints in how to proceed with this problem?

Answer 1

$ 3(3^{33})= 3^{x} $, then $ x = 34 $ .

Answer 2

$$a^b \times a^c = a^{(b+c)}. $$ I hope you got the answer as 34 now.

Answer 3

$3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3=3^x$

Answer 4

For the last line of your answer:

3(3³³) = 3^x

Apply the exponent rules for multiplication and you get:

3¹.3³³ = 3^x

3^(1+33) = 3^x

Then we got

1+33 = x

x = 34