Given that 4^x=8^2-x find the value of x?

Question

As above probably really easy but I can't do it.
Thanks matt

Answer 1

2^(2x)=2^(3(2-x))
2x=3(2-x)
2x=6-3x
5x=6
x=6/5

Answer 2

I assume you mean 4^x = 8^(2-x). Start by rewriting 4 and 8 as powers of 2. Then you get (2^2)^x = (2^3)^(2-x). The rule for exponents of exponents is that you multiply the exponents, so you get 2^(2x) = 2^(6-3x). That's just 2x = 6 - 3x, which you can solve algebraically. It's tricky if exponents are new to you, so don't worry about it!

Answer 3

4^x=8^(2-x)
=>2^2x=2^(6-3x)
if the base id the same the indices can be compared
so 2x=6-3x
adding 3x
5x=6
dividing by 5 x=6/5

Answer 4

I also did this problem and camp up with x=6/5. I checked with the calculator and they both equaled on both sides. Whoever gave the answerers a thumbs down for their right answer is jealous. Shame on that person for doing that.

Answer 5

u can ask at http://cbsemaths.co.in for prompt reply. The CBSEMaths Online Team will reply you soon

Is it 4^x=8^(2-x) or 4^x=8^2 - x ?

first one has very easy answer but second one need more mind practice

Answer 6

4^x=8(²-x)
Converting to the same base =>2²^x = 2(³.²^-x)
Equation expoent:
2x = 3.2-x
2x + x = 6
3x = 6
x = 6:3=2
Solution: {x E R | x = 2}
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Pls check result before to make a negative point. Tks + kisses.

Answer 7

2^2x=2^6-3x

2x=6-3x
x=6/5