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Maths question

Given that 4x=82-x, find the value of x. Without using logs!

Forgot how to do this simple question -_- (I know you have to write both numbers in the form of 2)
4^x=8^(2-x)
2^(2x)=2^(3(2-x))
2^(2x)=2^(6-3x)
2x = 6 - 3x
x=1.2
Reply 2
Original post by The Clockwork Apple
4^x=8^(2-x)
2^(2x)=2^(3(2-x))
2^(2x)=2^(6-3x)
2x = 6 - 3x
x=1.2


I actually just got it right after I posted this, but thank you anyway!
Original post by rm_27
I actually just got it right after I posted this, but thank you anyway!

That's OK :biggrin:
Original post by The Clockwork Apple

2^(2x)=2^(6-3x)
2x = 6 - 3x

oOHoh but u used logs there
Original post by CancerousProblem
oOHoh but u used logs there
No
Original post by The Clockwork Apple
No

Yes u did that's how u got from the line there to the line on the bottom u just didnt write it
Original post by CancerousProblem
Yes u did that's how u got from the line there to the line on the bottom u just didnt write it


a^b = a^c => b = c, no? That's just a property of exponentiation, is it not?
Original post by StrangeBanana
a^b = a^c => b = c, no?

yeah if u take logs then it is

f(a) = f(b) does not imply a = b, only the other way around.

you have to take logs
Original post by CancerousProblem
yeah if u take logs then it is


:| I don't think you need logs to prove that result
Original post by CancerousProblem
f(a) = f(b) does not imply a = b, only the other way around.

you have to take logs


It does if f is a bijective function like 2^x
(edited 8 years ago)
Original post by StrangeBanana
Not in general, it doesn't, but in this case it does

because the result holds for logarithims
Original post by CancerousProblem
because the result holds for logarithims


see 2 posts above, there is no need to use logarithms
(edited 8 years ago)

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