Question

hey dear

answer the question..​

Answer 1

Answer:

x = (log3/2)/log2

y = (log3/4)/log3

Step-by-step explanation:

Given,

$2^{x-1}=3^y\\\;\\\textbf{Taking log of both sides}\\\;\\(x-1)\log 2=y\log 3\;\;\;\;...............I)$

And,

$4^x=3^{y+1}\\\;\\\textbf{Taking log of both sides}\\\;\\x\log4=(y+1)\log3\\\;\\x\log2^2=y\log3+\log3\\\;\\2x\log2=y\log3+\log3\;\;\;\;\;\;.............II)$

Now, Subtracting equation I) from Equation 2,

$2x\log2-(x-1)\log2=\log3\\\;\\(2x-x+1)\log2=\log3\\\;\\(x+1)=\frac{\log3}{\log2}\\\;\\x=\frac{\log3}{\log2}-1\\\;\\x=\frac{\log3-\log2}{\log2}\\\;\\x=\frac{\log\frac{3}{2}}{\log2}$

Now,

From equation I)

$x\log2-\log2=y\log3\\\;\\\frac{\log\frac{3}{2}}{\log2}\log2-\log2=y\log3\\\;\\\log\frac{3}{2}-\log2=y\log3\\\;\\\log3-\log2-\log2=y\log3\\\;\\\log3-2\log2=y\log3\\\;\\\log3-\log2^2=y\log3\\\;\\\log3-\log4=y\log3\\\;\\y=\frac{\log3-\log4}{\log3}\\\;\\y=\frac{\log\frac{3}{4}}{\log3}$