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# 2^3x-1=(1\8)^x

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2^3x-1=(1\8)^x

Guest Mar 4, 2015

#2
+26642
+5

If this is

$$2^{3x-1}=(\frac{1}{8})^x$$

then rewrite it as

$$\\2^{3x-1}=(\frac{1}{2^3})^x\\\\2^{3x-1}=(2^{-3})^x\\\\2^{3x-1}=2^{-3x}$$

So we must have

$$\\3x-1=-3x\\\\6x=1\\\\x=\frac{1}{6}$$

.

Alan  Mar 4, 2015
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#1
+47
0

x=8^(-x-1)*((8^x)+1)

MILAN6030  Mar 4, 2015
#2
+26642
+5

If this is

$$2^{3x-1}=(\frac{1}{8})^x$$

then rewrite it as

$$\\2^{3x-1}=(\frac{1}{2^3})^x\\\\2^{3x-1}=(2^{-3})^x\\\\2^{3x-1}=2^{-3x}$$

So we must have

$$\\3x-1=-3x\\\\6x=1\\\\x=\frac{1}{6}$$

.

Alan  Mar 4, 2015

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