Answer:
[tex]x=\frac{5}{7}[/tex]
Step-by-step explanation:
[tex](4^{x-1}) (\frac{1}{2} )^{3-2x}=(\frac{1}{8} )^x[/tex] (I changed 0.5 to 1/2 for convenience sake)
[tex](2^{2(x-1)})(\frac{1}{2 })^{{3-2x}}=(\frac{1}{2})^{3x} }[/tex]
[tex](2^{2x-2})=\frac{(\frac{1}{2})^{3x} }{(\frac{1}{2 })^{{3-2x}}}[/tex]
[tex](\frac{1}{2})^{-(2x-2)= (\frac{1}{2}) ^{3x-(3-2x)}[/tex]
[tex](\frac{1}{2})^{2-2x}=(\frac{1}{2})^{5x-3}[/tex]
[tex]2-2x=5x-3[/tex]
[tex]5=7x[/tex]
I assumed you got the answer already as you asked for the process, so I will spare much explanations.
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Verified answer
Answer:
[tex]x=\frac{5}{7}[/tex]
Step-by-step explanation:
[tex](4^{x-1}) (\frac{1}{2} )^{3-2x}=(\frac{1}{8} )^x[/tex] (I changed 0.5 to 1/2 for convenience sake)
[tex](2^{2(x-1)})(\frac{1}{2 })^{{3-2x}}=(\frac{1}{2})^{3x} }[/tex]
[tex](2^{2x-2})=\frac{(\frac{1}{2})^{3x} }{(\frac{1}{2 })^{{3-2x}}}[/tex]
[tex](\frac{1}{2})^{-(2x-2)= (\frac{1}{2}) ^{3x-(3-2x)}[/tex]
[tex](\frac{1}{2})^{2-2x}=(\frac{1}{2})^{5x-3}[/tex]
[tex]2-2x=5x-3[/tex]
[tex]5=7x[/tex]
[tex]x=\frac{5}{7}[/tex]
I assumed you got the answer already as you asked for the process, so I will spare much explanations.