Answers

2015-02-22T23:57:51+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
8*2^{2x}+4*2^x*2^1=1+2^x\\\\Let\ 2^x=y,\ \ x=Log_2\ y\\\\8y^2+4*2y=1+y\\\\8y^2+7y-1=0\\\\8y^2+8y-y-1=0\\\\8y(y+1)-(y+1)=0\\\\(8y-1)(y+1)=0\\\\y=-1\ \ or\ \ 1/8\\\\y\ cannot\ be\ negative,\ as\ y=2^x\\\\y=\frac{1}{2^3}\\\\x=Log_2\ y=-Log_2\ 2^3=-3
1 5 1
2015-02-23T09:24:12+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
To solve this question, you need to know logarithms.
Taking log with base 2 on both sides, we get
log8 + 2xlog2 + log4 + (x+1)log2 = log1 + xlog2
3 + 2x + 2 + x + 1 = x
-2x = 6
x = -3
0