Question

if a^p=x a^q=y x^q.y^p=2ar then pqr=

Answer 1

Answer:

$\bold{p.q.r=log_a x.log_a y.a^{2log_ax.log_a y-log_a2-1}} \\ or \\ \bold{p.q.r=log_a x.log_a y.\frac{a^{2.log_a x.log_a y}}{2a}}$

Step-by-step explanation:

$\huge\mathcal{Given}</p><p>\\</p><p></p><p>a^p=x ==>p=log_a x ---->\bold{EQ\:01}</p><p>\\</p><p>a^q=y==>q=log_a y ---->\bold{EQ\:02}</p><p></p><p>\\</p><p>x^q.y^p=2a.r. ---->\bold{EQ\:03}</p><p>\\ \\ </p><p>From\:EQ\:03</p><p>\\ \\ </p><p></p><p>log_a(x^q.y^p)=log_a(2a.r)</p><p>\\</p><p>log_ax^q+log_ay^p=log_a2+log_aa+log_ar</p><p>\\</p><p></p><p> \\ </p><p></p><p>we\:know\:that\:</p><p>\\ \\ </p><p></p><p>\bold{log\:a+log\:b=log(a.b)}</p><p>\\</p><p>q.log_ax+p.log_ay=log_a2+1+log_ar</p><p>\\</p><p> \\ </p><p>from\:EQ\:01\:and\:02\:</p><p>\\</p><p> \\ </p><p>log_a y.log_ax+log_a x.log_ay=log_a2+1+log_ar</p><p>\\</p><p>2log_ax.log_a y=log_a2+1+log_ar</p><p>\\</p><p>log_ar=2log_ax.log_a y-log_a2-1</p><p>\\</p><p>r=a^{2log_ax.log_a y-log_a2-1}--->\bold{EQ\:04}</p><p>\\</p><p> \\ </p><p>Now\:Multiply \:Equations\:01,02\:and\:04</p><p>\\ \\ </p><p>\bold{p.q.r=log_a x.log_a y.a^{2log_ax.log_a y-log_a2-1}}----->\bold{EQ\:05}</p><p></p><p>\\ \\ </p><p>Or</p><p>\\ \\ </p><p>x^q.y^p=2a.r.</p><p>\\</p><p>r=\frac{x^q.y^p}{2a}</p><p>\\</p><p>r=\frac{(a^p)^q.(a^q)^p}{2a}</p><p>\\</p><p>r=\frac{(a^{pq})^2}{2.a}</p><p>\\</p><p>r=\frac{a^{2p.q}}{2a}</p><p>\\</p><p></p><p>r=\frac{a^{2.log_a x.log_a y}}{2a}</p><p>\\</p><p></p><p>\\</p><p>you \:can\: replace\: this \:r \:value\: in \:eq\: 05</p><p> \\ </p><p>$

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