Question

find x.
2^((x+6)/2)+2^(x+2)=5

Answer 1

x is -2

Answer 2

You can solve this by trial and error and some guess work.
2 is raised to some power and then added to another 2 raised to some power to give an odd value of 5. One possibility is 4 + 1 = 5 which implies x + 2 = 0 or x = -2
And if we put x = -2 in (x+6)/2 we get 2 and so 2^2 + 2^0 = 5

Answer 3

u can get an exact solution if u use integer method as i got

Answer 4

2^((x+6)/2 ) + 2^(x+2) = 5
2^(x/2 + 3) + 2^(x+2) = 5
2^x/2 * 2^3 + 2^x * 2^2 - 5 = 0
4(2^x) + 8(2^x/2) - 5 = 0
(2*2^x/2 -1 ) (2*2^x/2 + 5) = 0
take each factor equals zero, then solve for x

Answer 5

2*2^x/2 -1 = 0
2*2^x/2 = 1
2^x/2 = 1/2
2^x/2 = 2^-1
x/2 = -1
x = -2

Answer 6

\(\Large 2^{\frac{x+6}{2}}+2^{x+2}=5\)
\(\Large \Rightarrow 2^{\frac{x+2}{2}}\times 2^{\frac{4}{2}} +2^{2(\frac{x+2}{2})}=5 \)

Let \(\large t=2^{\frac{x+2}{2}}\), the equation become:
\(\Large 4t+t^{2}=5\\\Large \Rightarrow t^2+4t-5=0\\\Large \Rightarrow t=1\ \ or\ t=-5 \)

\(\Large {\color{red}{t=1}} \Rightarrow 2^{\frac{x+2}{2}}=1 \Rightarrow \frac{x+2}{2}=log_{2}{1}\)

\(\Large {\color{red}{t=-5}} \Rightarrow 2^{\frac{x+2}{2}}=-5\Rightarrow \frac{x+2}{2} =log_{2}{-5}\)

Answer 7

Last line: Can't take logarithm of a negative number.

Answer 8

@RadEn no, that is 4t because \(\Large \Rightarrow 2^{\frac{x+2}{2}}\times 2^{\frac{4}{2}} =2^2\times2^{\frac{x+2}{2}}=4t\)

Answer 9

@aum yup

Answer 10

so \[\log_{2}1=0 \]x+2=0
x=-2

Answer 11

yeah! great @amirreza1870 :)