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Question Number 191833 by Mingma last updated on 01/May/23

Answered by deleteduser1 last updated on 01/May/23

$$a={log}_{2}900;b={log}_{3}900;c={log}_{5}900$$$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}={log}_{900}2+{log}_{900}3+{log}_{900}5$$$$={log}_{900}\left(2\times 3\times 5\right)={log}_{900}\left(30\right)=\frac{1}{2}$$

Answered by BaliramKumar last updated on 01/May/23

$$2^a=900$$$$3^b=900$$$$5^c=900$$$$2^a\times 3^b\times 5^c=900\times 900\times 900={\left({30}^2\right)}^3$$$$2^a\times 3^b\times 5^c={30}^6={\left(2\times 3\times 5\right)}^6$$$$2^a\times 3^b\times 5^c=2^6\times 3^6\times 5^6$$$$a=b=c=6$$$$\frac{6\times 6+6\times 6+6\times 6}{6\times 6\times 6}=\frac{3\times \cancel{6}\times \cancel{6}}{6\times \cancel{6}\times \cancel{6}}=\frac{3}{6}=\frac{1}{2}$$$$\mathrm{answer}\mathrm{is}\mathrm{right}\mathrm{by}\mathrm{mistake}$$

Commented by Tinku Tara last updated on 01/May/23

$$2^6\ne 900$$$$2^a{3}^b{5}^c={30}^6\nRightarrow a=b=c=6$$

Commented by BaliramKumar last updated on 01/May/23

$$\mathrm{sorry}\mathrm{i}\mathrm{ignore}\mathrm{first}\mathrm{three}\mathrm{equ}.$$

Answered by HeferH last updated on 01/May/23

$$2^{\mathrm{a}}=900$$$$3^{\mathrm{b}}=900$$$$5^{\mathrm{c}}=900$$$$2\cdot 3\cdot 5={900}^{\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}}}$$$$30={30}^{2(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}})}$$$$\therefore \frac{\mathrm{ab}+\mathrm{bc}+\mathrm{ac}}{\mathrm{abc}}=\frac{1}{2}$$

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