Please Help

[hotvicky]

4^x=8^x+1

 

[Daniel_Feldman]

What do you need to do?

 

[hotvicky]

i need to solve without using logarithms

 

[Daniel_Feldman]

4^x=8^x+1

Are you sure it's not 8^(x+1)?

 

[hotvicky]

yeah it is 8^(x+1)

 

[Daniel_Feldman]

Oh yay...that makes it easier...otherwise we would have had ourselves an empty set.


4^x=8^(x+1)

Think about this:

If, 2^a=2^b, then a=b.

If 1000^(abcdefg)=1000^(hijkl), then abcdefg=hijkl

Makes sense? That means we should start by getting the same base on each side. In this case, two would make sense, as 2^2=4 and 2^3=8.


(2^2)^x=(2^3)^(x+1)

Now, I'm not sure if you're familiar with this rule of exponents, but it states that

(x^a)^b=x^(ab)


Thus, we rewrite as

2^(2x)=2^(3x+3)

Now, we use what I explained at the beginning:

2x=3x+3

so

x=-3


It is always good to check your answer by plugging it back into the original equation.

(4^x)=8^(x+1)

4^(-3)=1/(4^3)=1/64

8^(-3+1)=8^(-2)=1/(8^2)=1/64


The two sides are equal, so x=3 is confirmed to be correct, and we are done.


Hope this helps,

-Daniel-

 

[hotvicky]

Thank you so much now it is so much easier for me to do the next few.

 

[Daniel_Feldman]

You're quite welcome. Good luck.