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# Maths question watch

1. 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)
2. 4^x=8^(2-x)
2^(2x)=2^(3(2-x))
2^(2x)=2^(6-3x)
2x = 6 - 3x
x=1.2
3. (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!
4. (Original post by rm_27)
I actually just got it right after I posted this, but thank you anyway!
That's OK
5. (Original post by The Clockwork Apple)
2^(2x)=2^(6-3x)
2x = 6 - 3x
oOHoh but u used logs there
6. (Original post by CancerousProblem)
oOHoh but u used logs there
No
7. (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
8. (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?
9. (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
10. (Original post by CancerousProblem)
yeah if u take logs then it is
:| I don't think you need logs to prove that result
11. (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
12. (Original post by StrangeBanana)
Not in general, it doesn't, but in this case it does
because the result holds for logarithims
13. (Original post by CancerousProblem)
because the result holds for logarithims
see 2 posts above, there is no need to use logarithms

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