Need help with exponential equations

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

The solution is the following:

2^{x} + 2 * 2^{x}+ 4 * 2^{x} = 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 * 2^{x} = 56

2^{x} = 8

How did 56 turn into 8? Shouldn't it be 8^{2} instead of just 8?

2^{x} = 2^{3 }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.

Please reply,

thanks.

Re: Need help with exponential equations

We are given:

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

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

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

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

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

$\displaystyle 7\cdot2^x=56$

Divide through by 7:

$\displaystyle 2^x=8=2^3$

Hence:

$\displaystyle x=3$

Re: Need help with exponential equations

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

However, I don't understand this part:

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

Wouldn't

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

be 2^{x }+ 4^{x} + 8^{x }= 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

Re: Need help with exponential equations

Quote:

Originally Posted by

**dirtyharry** ...

I don't understand this part:

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

Wouldn't

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

be 2^{x }+ 4^{x} + 8^{x }= 56?

How did you come up with the "1"?

Note that:

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

is not the same as:

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

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

$\displaystyle u+2u+4u=56$

Now combine like terms on the left:

$\displaystyle 7u=56$

Divide both sides by 7:

$\displaystyle u=8$

Back-substitute for $\displaystyle u$:

$\displaystyle 2^x=8=2^3$

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

Re: Need help with exponential equations

Quote:

Originally Posted by

**MarkFL2** Note that:

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

is not the same as:

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

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

$\displaystyle u+2u+4u=56$

Now combine like terms on the left:

$\displaystyle 7u=56$

Divide both sides by 7:

$\displaystyle u=8$

Back-substitute for $\displaystyle u$:

$\displaystyle 2^x=8=2^3$

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

Thanks :D