Algebra Questions

Question Number 101498 by bemath last updated on 03/Jul/20

$$\mathrm{If}\:\mathrm{4}^{\mathrm{x}} \:=\:\mathrm{5}^{\mathrm{y}} \:=\:\mathrm{20}^{\mathrm{z}} \:. \\$$$$\mathrm{what}\:\mathrm{is}\:\mathrm{z}\:\mathrm{in}\:\mathrm{term}\:\mathrm{of}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}. \\$$$$\bigstar\spadesuit \\$$

Commented by bobhans last updated on 03/Jul/20

$$\mathrm{x}\:\mathrm{ln}\left(\mathrm{4}\right)\:=\:\mathrm{y}\:\mathrm{ln}\left(\mathrm{5}\right)\:=\:\mathrm{z}\:\left(\mathrm{ln}\left(\mathrm{4}\right)+\mathrm{ln}\left(\mathrm{5}\right)\right) \\$$$$\Rightarrow\left(\mathrm{1}\right)\:\mathrm{ln}\left(\mathrm{4}\right)\:=\:\frac{\mathrm{y}\:\mathrm{ln}\left(\mathrm{5}\right)}{\mathrm{x}} \\$$$$\Rightarrow\left(\mathrm{2}\right)\:\mathrm{z}\:=\:\frac{\mathrm{y}\:\mathrm{ln}\left(\mathrm{5}\right)}{\mathrm{ln}\left(\mathrm{4}\right)+\mathrm{ln}\left(\mathrm{5}\right)}\:=\:\frac{\mathrm{y}\:\mathrm{ln}\left(\mathrm{5}\right)}{\frac{\mathrm{y}\:\mathrm{ln}\left(\mathrm{5}\right)}{\mathrm{x}}\:+\:\mathrm{ln}\left(\mathrm{5}\right)} \\$$$$\:\:\:\:\:\mathrm{z}\:=\:\frac{\mathrm{yx}}{\mathrm{y}+\mathrm{x}}\:=\:\frac{{xy}}{{x}+{y}}\:\heartsuit\lozenge \\$$

Commented by Dwaipayan Shikari last updated on 03/Jul/20

$$\mathrm{4}^{{x}} =\mathrm{5}^{{y}} =\mathrm{20}^{{z}} ={k}\:\:\left({k}\neq\mathrm{0}\right) \\$$$$\mathrm{4}={k}^{\frac{\mathrm{1}}{{x}}} \:\:\:\:\:\mathrm{5}={k}^{\frac{\mathrm{1}}{{y}}} \:\:\:\:\:\mathrm{20}={k}^{\frac{\mathrm{1}}{{z}}} \\$$$${k}^{\frac{\mathrm{1}}{{x}}} {k}^{\frac{\mathrm{1}}{{y}}} ={k}^{\frac{\mathrm{1}}{{z}}} \\$$$$\frac{\mathrm{1}}{{z}}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\:\:\:\:\Rightarrow{z}=\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\right)^{−\mathrm{1}} =\frac{{xy}}{{x}+{y}}\:\:\blacksquare\:\bigstar\:\mathscr{L} \\$$

Commented by bramlex last updated on 03/Jul/20

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{both}\: \\$$

Answered by RKT last updated on 03/Jul/20

$$\mathrm{4}^{{x}} =\mathrm{20}^{{z}} \\$$$$\Rightarrow\mathrm{4}^{{x}} =\mathrm{4}^{{z}} ×\mathrm{5}^{{z}} \\$$$$\Rightarrow\mathrm{4}^{{x}−{z}} =\mathrm{5}^{{z}} \\$$$$\Rightarrow\mathrm{4}^{\frac{{x}−{z}}{{z}}} =\mathrm{5} \\$$$$\Rightarrow\mathrm{4}^{\frac{{y}}{{z}}\left({x}−{z}\right)} =\mathrm{5}^{{y}} \\$$$$\Rightarrow\mathrm{4}^{\frac{{y}}{{z}}\left({x}−{z}\right)} =\mathrm{4}^{{x}} \because\mathrm{4}^{{x}} =\mathrm{5}^{{y}} \\$$$$\Rightarrow\frac{{y}}{{z}}\left({x}−{z}\right)={x} \\$$$$\Rightarrow{xy}−{yz}={zx} \\$$$$\Rightarrow{xy}={zx}+{yz} \\$$$$\Rightarrow{xy}={z}\left({x}+{y}\right) \\$$$$\Rightarrow\frac{{xy}}{{x}+{y}}={z} \\$$