# Logarithm simplification (probably easy)

• Jan 21st 2011, 11:55 AM
Natalie11391
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 (Wondering) Thank you!
• Jan 21st 2011, 12:00 PM
dwsmith
Quote:

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 (Wondering) Thank you!

• Jan 21st 2011, 12:06 PM
Unknown008
You don't need to use logs... well not now.

Make a substitution:

$y = 2^x$

• Jan 21st 2011, 12:07 PM
e^(i*pi)
Quote:

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: $2^x > 0$
• Jan 21st 2011, 12:09 PM
Unknown008
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.
• Jan 21st 2011, 01:04 PM
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!
• Jan 21st 2011, 01:08 PM
e^(i*pi)
Quote:

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