# Thread: Logarithm simplification (probably easy)

1. ## Logarithm simplification (probably easy)

Ok so as the title says, it's probably really easy but I'm having trouble getting a value of x:

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

I've got down to

xlog(4) = log(2^(x+2)-3)

And I have NO idea how to simplify to get rid of that -3.... Please help! I'm sure there's something REALLY obvious that I'm missing Thank you!

2. Originally Posted by Natalie11391
Ok so as the title says, it's probably really easy but I'm having trouble getting a value of x:

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

I've got down to

xlog(4) = log(2^(x+2)-3)

And I have NO idea how to simplify to get rid of that -3.... Please help! I'm sure there's something REALLY obvious that I'm missing Thank you!

3. You don't need to use logs... well not now.

Make a substitution:

$y = 2^x$

4. Originally Posted by Unknown008
You don't need to use logs... well not now.

Make a substitution:

$y = 2^x$

You wouldn't believe how hard I'm kicking myself now.

To the OP, when working out your solutions remember that for all real x: $2^x > 0$

5. Oh, I think I know how you feel, I felt the same thing before getting the idea for the substitution.

Note to OP: Sometimes, taking logs immediately only make things harder. Look at the bases first.

6. Thanks for your help, but I still can't do it!! It must be lack of sleep but I can't substitute 2^x as y for 4^x

I know that 2^2 = 4 so 2^2x = 4^x but then I can't seem to get my head around what to do next!

7. Originally Posted by Natalie11391
Thanks for your help, but I still can't do it!! It must be lack of sleep but I can't substitute 2^x as y for 4^x

I know that 2^2 = 4 so 2^2x = 4^x but then I can't seem to get my head around what to do next!
That's right so sub in $2^{2x}$ for 4^x

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

$2^{2x} +3 = 2^{x+2}$

You also know that $2^{x+2} = 4 \cdot 2^x$ (which is the reverse of what you did in the OP)

$2^{2x} + 3 = 4 \cdot 2^x$ and if you rearrange it you get $(2^x)^2 - 4 \cdot 2^x + 3 = 0$

That is a quadratic equation in $2^x$ so solve using your favourite method