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Question Number 120120 by benjo_mathlover last updated on 29/Oct/20

$$\{\begin{array}{l}\log {}_a\left(x\right)=8\\ \log {}_b\left(x\right)=3\\ \log {}_c\left(x\right)=6\end{array}\Rightarrow \log {}_{abc}\left(x\right)=?$$

Answered by bemath last updated on 29/Oct/20

$$\{\begin{array}{l}\log {}_a\left(x\right)=8\Rightarrow \log {}_x\left(a\right)=\frac{1}{8}\\ \log {}_b\left(x\right)=3\Rightarrow \log {}_x\left(b\right)=\frac{1}{3}\\ \log {}_c\left(x\right)=6\Rightarrow \log {}_x\left(c\right)=\frac{1}{6}\end{array}$$$$Then\log {}_{abc}\left(x\right)=\frac{1}{\log {}_x\left(abc\right)}=\frac{1}{\frac{1}{8}+\frac{1}{3}+\frac{1}{6}}$$$$=\frac{24}{3+8+4}=\frac{24}{15}=\frac{8}{5}$$

Answered by $@y@m last updated on 29/Oct/20

$$x=a^8=b^3=c^6$$$$a=x^{\frac{1}{8}},b=x^{\frac{1}{3}},c=x^{\frac{1}{6}}$$$$abc=x^{\frac{1}{8}+\frac{1}{3}+\frac{1}{6}}=x^{\frac{15}{24}}$$$${\log }_x\left(abc\right)=\frac{15}{24}$$$$\log {}_{abc}\left(x\right)=\frac{24}{15}=\frac{8}{5}$$

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