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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!
Already answered belowQuote:
Originally Posted by Natalie11391
You don't need to use logs... well not now.
Make a substitution:
$\displaystyle y = 2^x$
How about it? (Smile)
You wouldn't believe how hard I'm kicking myself now.Quote:
Originally Posted by Unknown008
To the OP, when working out your solutions remember that for all real x: $\displaystyle 2^x > 0$
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.
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 $\displaystyle 2^{2x}$ for 4^xQuote:
Originally Posted by Natalie11391
(4^(x)+3)/2 = 2^(x+1)
$\displaystyle 2^{2x} +3 = 2^{x+2}$
You also know that $\displaystyle 2^{x+2} = 4 \cdot 2^x$ (which is the reverse of what you did in the OP)
$\displaystyle 2^{2x} + 3 = 4 \cdot 2^x$ and if you rearrange it you get $\displaystyle (2^x)^2 - 4 \cdot 2^x + 3 = 0$
That is a quadratic equation in $\displaystyle 2^x$ so solve using your favourite method
