# Thread: Need help with exponential equations

1. ## Need help with exponential equations

2x + 2x+1 + 2x+2 = 56

The solution is the following:

2x + 2 * 2x+ 4 * 2x = 56

Why is that? I don't understand how it came to this. Where did the 4 come from? How come there's so many 2s?

7 * 2x = 56

2x = 8

How did 56 turn into 8? Shouldn't it be 82 instead of just 8?

2x = 23

x = 3

I only understand the last two steps. Please help, I need to know this (and more) for a test. This isn't the only equation that confuses me, but it would help to know what lead to the final solution.
thanks.

2. ## Re: Need help with exponential equations

We are given:

$2^x+2^{x+1}+2^{x+2}=56$

Using the property of exponents $a^b\cdot a^c=a^{b+c}$ we may rewrite the equation as:

$2^x+2^1\cdot2^x+2^2\cdot2^x=56$

$2^x+2\cdot2^x+4\cdot2^x=56$

$2^x(1+2+4)=56$

$7\cdot2^x=56$

Divide through by 7:

$2^x=8=2^3$

Hence:

$x=3$

3. ## Re: Need help with exponential equations

Thanks, now I know how the conversion to multiplying occured.

However, I don't understand this part:

$2^x(1+2+4)=56$

Wouldn't

$2^x+2\cdot2^x+4\cdot2^x=56$

be 2x + 4x + 8x = 56?

How did you come up with the "1"?

Also, about dividing through by 7: 7 is not their common denominator - 2 is not in seven (at least not without a decimal). How does it stay unchanged? Is it because of the X?

Thanks

4. ## Re: Need help with exponential equations

Originally Posted by dirtyharry
...

I don't understand this part:

$2^x(1+2+4)=56$

Wouldn't

$2^x+2\cdot2^x+4\cdot2^x=56$

be 2x + 4x + 8x = 56?

How did you come up with the "1"?
Note that:

$2^x+2\cdot2^x+4\cdot2^x=56$

is not the same as:

$2^x+(2\cdot2)^x+(4\cdot2)^x=56$

To make things simpler to see, let $u=2^x$ then we have:

$u+2u+4u=56$

Now combine like terms on the left:

$7u=56$

Divide both sides by 7:

$u=8$

Back-substitute for $u$:

$2^x=8=2^3$

You see, when I factored in my first post, I used the fact that each term on the left has $2^x$ as a factor.

5. ## Re: Need help with exponential equations

Originally Posted by MarkFL2
Note that:

$2^x+2\cdot2^x+4\cdot2^x=56$

is not the same as:

$2^x+(2\cdot2)^x+(4\cdot2)^x=56$

To make things simpler to see, let $u=2^x$ then we have:

$u+2u+4u=56$

Now combine like terms on the left:

$7u=56$

Divide both sides by 7:

$u=8$

Back-substitute for $u$:

$2^x=8=2^3$

You see, when I factored in my first post, I used the fact that each term on the left has $2^x$ as a factor.
Thanks