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

If 4^x = 5^y = 20^z . what is z in term of x and y. ★♠

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

Commented by bobhans last updated on 03/Jul/20

x ln(4) = y ln(5) = z (ln(4)+ln(5)) ⇒(1) ln(4) = ((y ln(5))/x) ⇒(2) z = ((y ln(5))/(ln(4)+ln(5))) = ((y ln(5))/(((y ln(5))/x) + ln(5))) z = ((yx)/(y+x)) = ((xy)/(x+y)) ♥◊

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

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

4^x =5^y =20^z =k (k≠0) 4=k^(1/x) 5=k^(1/y) 20=k^(1/z) k^(1/x) k^(1/y) =k^(1/z) (1/z)=(1/x)+(1/y) ⇒z=((1/x)+(1/y))^(−1) =((xy)/(x+y)) ■ ★ L

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

Commented by bramlex last updated on 03/Jul/20

thank you both

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

Answered by RKT last updated on 03/Jul/20

4^x =20^z ⇒4^x =4^z ×5^z ⇒4^(x−z) =5^z ⇒4^((x−z)/z) =5 ⇒4^((y/z)(x−z)) =5^y ⇒4^((y/z)(x−z)) =4^x ∵4^x =5^y ⇒(y/z)(x−z)=x ⇒xy−yz=zx ⇒xy=zx+yz ⇒xy=z(x+y) ⇒((xy)/(x+y))=z

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

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