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Question Number 42670 by Tawa1 last updated on 31/Aug/18

$$\mathrm{If}\mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c}\mathrm{are}\mathrm{in}\mathrm{a}\mathrm{GP}.\mathrm{Prove}\mathrm{that}{\log }_{\mathrm{n}}\mathrm{a},{\log }_{\mathrm{n}}\mathrm{b},{\log }_{\mathrm{n}}\mathrm{c}\mathrm{are}\mathrm{in}\mathrm{AP}$$

Commented by maxmathsup by imad last updated on 31/Aug/18

$$wehave\frac{b}{a}=\frac{c}{b}\Rightarrow b^2=ac\Rightarrow 2ln\left(b\right)=ln\left(a\right)+ln\left(c\right)\Rightarrow$$$$\frac{2ln\left(b\right)}{ln\left(n\right)}=\frac{ln\left(a\right)}{ln\left(n\right)}+\frac{ln\left(c\right)}{ln\left(n\right)}\Rightarrow 2{log}_n\left(b\right)={log}_n\left(a\right)+{log}_n\left(c\right)\Rightarrow$$$${log}_n\left(b\right)-{log}_n\left(a\right)={log}_n\left(c\right)-log\left(b\right)(=r)\Rightarrow {log}_n\left(a\right),{log}_n\left(b\right){andlog}_n\left(c\right)arein$$$$AP.$$

Commented by Tawa1 last updated on 31/Aug/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by prof Abdo imad last updated on 31/Aug/18

$$thankyousir.$$

Answered by Rio Michael last updated on 31/Aug/18

$$a,b,cinGP$$$$b^2=ac$$$$taking{log}_{n}onbothsides$$$${log}_n{b}^2={log}_{n}ac$$$$2{log}_{n}b={log}_{n}a+{log}_{n}c$$$${log}_{n}b=\frac{{log}_{n}a+{log}_{n}c}{2}$$$$whichistheArithmeticmean\left(AM\right)[b=\frac{a+c}{2}]forA.Ps$$

Commented by Tawa1 last updated on 31/Aug/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Rio Michael last updated on 31/Aug/18

$$welcome$$

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