5-log-5-3-5-3-log-5-3-3-

Question Number 145119 by imjagoll last updated on 02/Jul/21

\frac{5^{\log_{\frac{5}{3}} (5)}}{3^{\log_{\frac{5}{3}} (3)}} = ?

Answered by Rasheed.Sindhi last updated on 03/Jul/21

\frac{5^{\log_{\frac{5}{3}} (5)}}{3^{\log_{\frac{5}{3}} (3)}} = ?

\log_{\frac{5}{3}} (5) = a; \log_{\frac{5}{3}} (3) = b

{(\frac{5}{3})}^{a} = 5; {(\frac{5}{3})}^{b} = 3

\frac{{(\frac{5}{3})}^{a}}{{(\frac{5}{3})}^{b}} = \frac{5}{3}

{(\frac{5}{3})}^{a - b} = {(\frac{5}{3})}^{1}

a - b = 1 \to b = a - 1

\frac{5^{\log_{\frac{5}{3}} (5)}}{3^{\log_{\frac{5}{3}} (3)}} = \frac{5^{a}}{3^{b}} = \frac{5^{a}}{3^{a - 1}}

= 3 {(\frac{5}{3})}^{a}

= 3 {(\frac{5}{3})}^{\log_{\frac{5}{3}} (5)} = 3 (5) = 15

\begin{matrix} b^{\log_{b} (a)} = a \end{matrix}

Answered by liberty last updated on 03/Jul/21

let {\begin{cases} \log_{\frac{5}{3}} (5) = x \Rightarrow 5^{x - 1} = 3^{x} \\ \log_{\frac{5}{3}} (3) = y \Rightarrow 5^{y} = 3^{y + 1} \end{cases}

{\begin{cases} 5 = 3^{\frac{x}{x - 1}} \\ {(3^{\frac{x}{x - 1}})}^{y} = 3^{y + 1} \end{cases} \Rightarrow \frac{x y}{x - 1} = y + 1

\Rightarrow x y = x y + x - y - 1

\Rightarrow x = y + 1 . s o t h e v a l u e o f

\frac{5^{\log_{\frac{5}{3}} (5)}}{3^{\log_{\frac{5}{3}} (3)}} = \frac{5^{x}}{3^{y}} = \frac{5 . 5^{y}}{3^{y}}

\Rightarrow 5 \times {(\frac{5}{3})}^{y} = 5 \times 3 = 15

Answered by Rasheed.Sindhi last updated on 03/Jul/21

\frac{5^{\log_{\frac{5}{3}} (5)}}{3^{\log_{\frac{5}{3}} (3)}}

= \frac{\frac{5^{\log_{\frac{5}{3}} (5)}}{3^{\log_{\frac{5}{3}} (5)}} \times 3^{\log_{\frac{5}{3}} (5)}}{3^{\log_{\frac{5}{3}} (3)}}

= {(\frac{5}{3})}^{\log_{\frac{5}{3}} (5)} \times 3^{\log_{\frac{5}{3}} (5) - \log_{\frac{5}{3}} (3)}

\begin{matrix} b^{\log_{b} a} = a \end{matrix}

= 5 \times 3^{\log_{\frac{5}{3}} (\frac{5}{3})} = 5 \times 3^{1} = 15

\begin{matrix} \log_{a} a = 1 \end{matrix}