tim x
1/8*16^x=2^x
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TH1: TH2:
-8(x-3)=24-16:2 -8(-x+3)=24-16:2
-8x+24=16 8x-24=16
-8x=16-24 8x=16+24
-8x=-8 8x=40
x=1 x=5
Vậy \(x\in1;5\)
-8/x-3/=24-16:2 Suy ra x-3=-8;8
-8/x-3/=24-8 x=11;-5
-8/x-3/=16
/x-3/=16+(-8)
/x-3/=8
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(2\times A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2\times A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\)
\(A=1-\frac{1}{128}\)
\(A=\frac{127}{128}\)
\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(2\times B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)
\(B=1-\frac{1}{16}=\frac{15}{16}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\Leftrightarrow4\times x+\frac{15}{16}=1\)
\(\Leftrightarrow4\times x=\frac{1}{16}\)
\(\Leftrightarrow x=\frac{1}{64}\)
x . 4 + 15/16 = 1
x . 4 = 1 – 15/16 = 1/16
x = 1/16 : 4 = 1/64
1/8*16^x=2^x
1/8=2^x/16^x
(2/16)^x=1/8
(1/8)^x=1/8
suy ra x=1
Vậy x=1
\(\frac{1}{8}\cdot16^x=2^x\)
\(\Leftrightarrow\frac{1}{8}=\frac{2^x}{16^x}\)
\(\Leftrightarrow\left(\frac{2}{16}\right)^x=\frac{1}{8}\)
\(\Leftrightarrow\left(\frac{1}{8}\right)^x=\frac{1}{8}\)
Suy ra x=1