Problem/Riddle: Algebra related!?

Question

I have this problem and I was wondering if somebody could help me solve it.

If 2^x + 2^x + 2^x + 2^x =2^k
Express 'x' in terms of 'k'

Thanks.

Answer 1

2^x+2^x+2^x+2^x=2^k

Restating the same as

4 * (2^x) = 2^k

2^2 * 2^x = 2^k

2^(2+x)=2^k

(2+x) = k

Therefore,
x= (k-2)

Answer 2

well if that is exponents then k equals the number of x's see, 2^2+2^2+2^2+2^2=2^8 u see?

Answer 3

Translate on the form of power:
2^x+2^x+2^x+2^x =(2^x)^4
(2^x)^4=2^k
let's take ln of both sides:
ln(2^x)^4 = ln2^k
we know: lna^n=nlna
x4ln2=kln2
x4=kln2/ln2
let's simplify by ln2:
x4=k or x=1/4k or
x=k/4
good luck

Answer 4

4(2^x)=2^k
log base 2 of 4(2^x)=k

Answer 5

2^x + 2^x + 2^x + 2^x = 2^k
4(2^x) = 2^k
log(base 2) [4(2^x)] = log(base 2) (2^k)
log(base 2) 4 + x log(base 2) 2 = k log(base 2) 2
2 + x = k
x = k - 2

I see now I did it the hard way. Susrsu did it the easy way.

Answer 6

2^x +2^x +2^x +2^x = 4(2^x) = 2^k

.: 2^x =(2^k)/4 Now use logarithms.

xlog2 = klog2 - log4

x = k log2/log2 - log4/log2

x = k - log4/log2