Question and Answers Forum

All Questions      Topic List

Logarithms Questions

Previous in All Question      Next in All Question      

Previous in Logarithms      Next in Logarithms      

Question Number 187406 by Fridunatjan08 last updated on 16/Feb/23

3^x .8^(x/(x+2)) =6ln0−log10

$$\mathrm{3}^{{x}} .\mathrm{8}^{\frac{{x}}{{x}+\mathrm{2}}} =\mathrm{6}{ln}\mathrm{0}−{log}\mathrm{10} \\ $$

Commented by Fridunatjan08 last updated on 16/Feb/23

can someone please help me?

$${can}\:{someone}\:{please}\:{help}\:{me}? \\ $$

Commented by Frix last updated on 17/Feb/23

ln 0 is not defined

$$\mathrm{ln}\:\mathrm{0}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined} \\ $$

Answered by a.lgnaoui last updated on 17/Feb/23

Ithink that you want to tel 6ln10  we suppose that thrue  log10=((ln10)/(ln10))=1  3^x .8^(x/(x+2)) =6ln10−1  a=b  (a>0 b>0⇒lna=lnb  xln3+(x/(x+2))ln8=ln[6ln10−1]  posons   k=ln[6ln10−1]  x(x+2)ln3+xln8=(x+2)k  x^2 ln3+2xln3+xln8=(x+2)k  x^2 ln3+x(2ln3+ln8−k)−2k=0    x^2 +(1/(ln3))(2ln3+3ln2−k)x−((2k)/(ln3))=0  [x+(1+((3ln2−k)/(2ln3)))]^2 −  [1+((3ln2−k)/(2ln3)))^2 +((2k)/(ln3))]=0    x+(1+((3ln2−k)/(2ln3)))=  (√((1+((3ln2−k)/(2ln3)))^2 +((2k)/(ln3))))    x=±(√((1+((3ln2−k)/(2ln3)))^2 +((2k)/(ln3)))) −1−((3kn2−k)/(2ln3))

$${Ithink}\:{that}\:{you}\:{want}\:{to}\:{tel}\:\mathrm{6}{ln}\mathrm{10} \\ $$$${we}\:{suppose}\:{that}\:{thrue} \\ $$$${log}\mathrm{10}=\frac{{ln}\mathrm{10}}{{ln}\mathrm{10}}=\mathrm{1} \\ $$$$\mathrm{3}^{{x}} .\mathrm{8}^{\frac{{x}}{{x}+\mathrm{2}}} =\mathrm{6}{ln}\mathrm{10}−\mathrm{1} \\ $$$${a}={b}\:\:\left({a}>\mathrm{0}\:{b}>\mathrm{0}\Rightarrow{lna}={lnb}\right. \\ $$$${xln}\mathrm{3}+\frac{{x}}{{x}+\mathrm{2}}{ln}\mathrm{8}={ln}\left[\mathrm{6}{ln}\mathrm{10}−\mathrm{1}\right] \\ $$$${posons}\:\:\:{k}={ln}\left[\mathrm{6}{ln}\mathrm{10}−\mathrm{1}\right] \\ $$$${x}\left({x}+\mathrm{2}\right){ln}\mathrm{3}+{xln}\mathrm{8}=\left({x}+\mathrm{2}\right){k} \\ $$$${x}^{\mathrm{2}} {ln}\mathrm{3}+\mathrm{2}{xln}\mathrm{3}+{xln}\mathrm{8}=\left({x}+\mathrm{2}\right){k} \\ $$$${x}^{\mathrm{2}} {ln}\mathrm{3}+{x}\left(\mathrm{2}{ln}\mathrm{3}+{ln}\mathrm{8}−{k}\right)−\mathrm{2}{k}=\mathrm{0} \\ $$$$ \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} +\frac{\mathrm{1}}{{ln}\mathrm{3}}\left(\mathrm{2}{ln}\mathrm{3}+\mathrm{3}{ln}\mathrm{2}−{k}\right)\boldsymbol{{x}}−\frac{\mathrm{2}{k}}{{ln}\mathrm{3}}=\mathrm{0} \\ $$$$\left[\boldsymbol{{x}}+\left(\mathrm{1}+\frac{\mathrm{3}{ln}\mathrm{2}−{k}}{\mathrm{2}{ln}\mathrm{3}}\right)\right]^{\mathrm{2}} − \\ $$$$\left.\left[\mathrm{1}+\frac{\mathrm{3}{ln}\mathrm{2}−{k}}{\mathrm{2}{ln}\mathrm{3}}\right)^{\mathrm{2}} +\frac{\mathrm{2}{k}}{{ln}\mathrm{3}}\right]=\mathrm{0} \\ $$$$ \\ $$$${x}+\left(\mathrm{1}+\frac{\mathrm{3}{ln}\mathrm{2}−{k}}{\mathrm{2}{ln}\mathrm{3}}\right)= \\ $$$$\sqrt{\left(\mathrm{1}+\frac{\mathrm{3}{ln}\mathrm{2}−{k}}{\mathrm{2}{ln}\mathrm{3}}\right)^{\mathrm{2}} +\frac{\mathrm{2}{k}}{{ln}\mathrm{3}}} \\ $$$$ \\ $$$$\boldsymbol{{x}}=\pm\sqrt{\left(\mathrm{1}+\frac{\mathrm{3}{ln}\mathrm{2}−{k}}{\mathrm{2}{ln}\mathrm{3}}\right)^{\mathrm{2}} +\frac{\mathrm{2}{k}}{{ln}\mathrm{3}}}\:−\mathrm{1}−\frac{\mathrm{3}{kn}\mathrm{2}−{k}}{\mathrm{2}{ln}\mathrm{3}} \\ $$

Commented by Fridunatjan08 last updated on 17/Feb/23

thsnks you bro.

$${thsnks}\:{you}\:{bro}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com