Ifa = log8225andb = log215,thenaintermsofbis−Ifa = log8225andb = log215,thenaintermsofbis−\eqalign{ & {\text{If}}\,{\text{a = lo}}{{\text{g}}_{\text{8}}}{\text{225}}\,{\text{and}}\,{\text{b = lo}}{{\text{g}}_{\text{2}}}{\text{15}}, \cr & {\text{then}}\,a\,{\text{in}}\,{\text{terms}}\,{\text{of}}\,b\,{\text{is}} - \cr} " /> Ifa = log8225andb = log215,thenaintermsofbis−Ifa = log8225andb = log215,thenaintermsofbis−\eqalign{ & {\text{If}}\,{\text{a = lo}}{{\text{g}}_{\text{8}}}{\text{225}}\,{\text{and}}\,{\text{b = lo}}{{\text{g}}_{\text{2}}}{\text{15}}, \cr & {\text{then}}\,a\,{\text{in}}\,{\text{terms}}\,{\text{of}}\,b\,{\text{is}} - \cr} " /> Ifa = log8225andb = log215,thenaintermsofbis−Ifa = log8225andb = log215,thenaintermsofbis−\eqalign{ & {\text{If}}\,{\text{a = lo}}{{\text{g}}_{\text{8}}}{\text{225}}\,{\text{and}}\,{\text{b = lo}}{{\text{g}}_{\text{2}}}{\text{15}}, \cr & {\text{then}}\,a\,{\text{in}}\,{\text{terms}}\,{\text{of}}\,b\,{\text{is}} - \cr} " />

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$\begin{aligned} If a = lo g_{8} 225 and b = lo g_{2} 15, \\ then a in terms of b is - \end{aligned}$ $\eqalign{ & {\text{If}}\,{\text{a = lo}}{{\text{g}}_{\text{8}}}{\text{225}}\,{\text{and}}\,{\text{b = lo}}{{\text{g}}_{\text{2}}}{\text{15}}, \cr & {\text{then}}\,a\,{\text{in}}\,{\text{terms}}\,{\text{of}}\,b\,{\text{is}} - \cr}$

1^b/₂
2^2b/₃
3b
4^3b/₂

Answer:- 1

Explanation:-

Solution:

\begin{aligned} a = \log_{8} 225 \\ = \log_{2^{3}} (15^{2}) \\ = \frac{2}{3} \log_{2} 15 \\ = \frac{2 b}{3} \end{aligned}

$\eqalign{ & a = {\log _8}225 \cr & = {\log _{{2^3}}}\left( {{{15}^2}} \right) \cr & = \frac{2}{3}{\log _2}15 \cr & = \frac{{2b}}{3} \cr}$

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", "text": "

\begin{aligned} If a = lo g_{8} 225 and b = lo g_{2} 15, \\ then a in terms of b is - \end{aligned}

$\eqalign{ & {\text{If}}\,{\text{a = lo}}{{\text{g}}_{\text{8}}}{\text{225}}\,{\text{and}}\,{\text{b = lo}}{{\text{g}}_{\text{2}}}{\text{15}}, \cr & {\text{then}}\,a\,{\text{in}}\,{\text{terms}}\,{\text{of}}\,b\,{\text{is}} - \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} a = \log_{8} 225 \\ = \log_{2^{3}} (15^{2}) \\ = \frac{2}{3} \log_{2} 15 \\ = \frac{2 b}{3} \end{aligned}

$\eqalign{ & a = {\log _8}225 \cr & = {\log _{{2^3}}}\left( {{{15}^2}} \right) \cr & = \frac{2}{3}{\log _2}15 \cr & = \frac{{2b}}{3} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} a = \log_{8} 225 \\ = \log_{2^{3}} (15^{2}) \\ = \frac{2}{3} \log_{2} 15 \\ = \frac{2 b}{3} \end{aligned}

$\eqalign{ & a = {\log _8}225 \cr & = {\log _{{2^3}}}\left( {{{15}^2}} \right) \cr & = \frac{2}{3}{\log _2}15 \cr & = \frac{{2b}}{3} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }