Question

Had me baffled for a while

2^x + 2^(4-x) = 17

Answer 1

\[2^x+\frac{2^4}{2^x}=17\] then multiply by
\[2^x\] and get a quadratic equation i think

Answer 2

I am getting x=0,4

Answer 3

Go on then...

Answer 4

I know u can "see2 it's 0 and 4, what are the steps...

Answer 5

\[2^x = a\]

Answer 6

and Now try solving quadratic way

Answer 7

I eventually hit on2^x + 2^(x-4) = 2^4 + 2^0

but I'm not convinced...

Answer 8

\[a^2 - 17a +16=0\]

Answer 9

get
\[2^{2x}+16=17\times 2^x\] or
\[2^{2x}-17\times 2^x+16=0\] a quadratic in
\[2^x\] solve
\[z^2-17z+16=0\] get
\[z=1, z = 16\] so
\[2^x=1\iff x=0\] or
\[2^x=16\iff x=4\]

Answer 10

TV time, just thought it was amusing and no-one was answering it....