# Finding x in a sum of exponents of the same base

This one has me stumped. I know that there is no law of exponents explaining what happens for a sum of exponents with the same base, so I tried taking the natural log on both sides and adding the exponents that way as they multiply the same base, but it doesn't seem to be correct and I haven't been able to find a good explanation for how to proceed.

The problem is: 2^(x+2) + 2^(x+1) + 2^(x) = 3/4

If this question is against any of SE's rules I would be more than pleased to know of a good source to learn how to solve these types of problems, or even simply a nudge in the right direction! Thanks.

write $$2^x\cdot 2^2+2^x\cdot 2+2^x=\frac{3}{4}$$ can you finish?

• The RHS is $3/4$, correct? – Tim Thayer Mar 20 '18 at 18:01
• yes this was a typo – Dr. Sonnhard Graubner Mar 20 '18 at 18:02
• I deeply appreciate your help! Thanks for assisting me in fostering greater mathematical intuition – Jake Mar 20 '18 at 18:39

$$2^{x+2} + 2^{x+1} + 2^x = 3/4$$

$$4 (2^x) +2(2^x) + 2^x = 3/4$$

Let $y=2^x$ and substitute to get $$7y=3/4$$

$$y= \frac{3}{28}$$

$$2^x= \frac{3}{28}$$

$$x \ln(2) = \ln(\frac{3}{28})$$

Solve for $x$.

• Thank you, my final answer of -3.222 coincides with the result I was able to reach on a website but without understanding the process. Very appreciative. – Jake Mar 20 '18 at 18:41
• Which part you did not get? – Mohammad Riazi-Kermani Mar 20 '18 at 18:46
• I meant before understanding the process with the help of your answer, sorry if I was unclear. – Jake Mar 21 '18 at 17:11
• No problem. Thanks for the comment. – Mohammad Riazi-Kermani Mar 21 '18 at 17:17