Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 1354 by Rasheed Ahmad last updated on 25/Jul/15

Slightly modified form of Q 1343. 3^(log(3x+4)) =4^(log(4x+3)) ,solve for x.

$${Slightly}\:{modified}\:{form}\:{of}\:{Q}\:\mathrm{1343}. \\ $$$$\mathrm{3}^{{log}\left(\mathrm{3}{x}+\mathrm{4}\right)} =\mathrm{4}^{{log}\left(\mathrm{4}{x}+\mathrm{3}\right)} ,{solve}\:{for}\:{x}. \\ $$

Commented by prakash jain last updated on 25/Jul/15

One option to prove that x=0 is the only solution for ((log (3x+4))/(log (4x+3)))−((log 4)/(log 3))=0 is to take derivative of the function. Also if the function has more than one zero it must change sign twice.

$$\mathrm{One}\:\mathrm{option}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that}\:{x}=\mathrm{0}\:\mathrm{is}\:\mathrm{the}\:\mathrm{only} \\ $$$$\mathrm{solution}\:\mathrm{for}\:\frac{\mathrm{log}\:\left(\mathrm{3}{x}+\mathrm{4}\right)}{\mathrm{log}\:\left(\mathrm{4}{x}+\mathrm{3}\right)}−\frac{\mathrm{log}\:\mathrm{4}}{\mathrm{log}\:\mathrm{3}}=\mathrm{0}\:\mathrm{is}\:\mathrm{to} \\ $$$$\mathrm{take}\:\mathrm{derivative}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}. \\ $$$$\mathrm{Also}\:\mathrm{if}\:\mathrm{the}\:\mathrm{function}\:\mathrm{has}\:\mathrm{more}\:\mathrm{than}\:\mathrm{one}\:\mathrm{zero} \\ $$$$\mathrm{it}\:\mathrm{must}\:\mathrm{change}\:\mathrm{sign}\:\mathrm{twice}. \\ $$

Answered by Yugi last updated on 25/Jul/15

3^(log(3x+4)) =4^(log(4x+3)) .........(1) Taking logs to base 10 on both sides of (1) [log(3x+4)]log3=[log(4x+3)]log4 ⇒((log(3x+4))/(log(4x+3)))=((log4)/(log3))...........(2) Upon observation of theform of (2), the only real solution is x=0 . A graphical approach yields the same result since (1) is equivalent to y=0 for y=3^(log(3x+4)) −4^(log(4x+3)) defined for x>−(3/4). Numerical methods work as well. Otherwise, I could not find an algebraic means of solving for x of the form ((log(ax+b))/(log(cx+d)))=e where all of a,b,c,d and e are non−zero reals.

$$\mathrm{3}^{{log}\left(\mathrm{3}{x}+\mathrm{4}\right)} =\mathrm{4}^{{log}\left(\mathrm{4}{x}+\mathrm{3}\right)} .........\left(\mathrm{1}\right) \\ $$$${Taking}\:{logs}\:{to}\:{base}\:\mathrm{10}\:{on}\:{both}\:{sides}\:{of}\:\left(\mathrm{1}\right) \\ $$$$\left[{log}\left(\mathrm{3}{x}+\mathrm{4}\right)\right]{log}\mathrm{3}=\left[{log}\left(\mathrm{4}{x}+\mathrm{3}\right)\right]{log}\mathrm{4} \\ $$$$\Rightarrow\frac{{log}\left(\mathrm{3}{x}+\mathrm{4}\right)}{{log}\left(\mathrm{4}{x}+\mathrm{3}\right)}=\frac{{log}\mathrm{4}}{{log}\mathrm{3}}...........\left(\mathrm{2}\right) \\ $$$${Upon}\:{observation}\:{of}\:{theform}\:{of}\:\left(\mathrm{2}\right),\:{the}\:{only}\:{real}\: \\ $$$${solution}\:{is}\:{x}=\mathrm{0}\:.\:{A}\:{graphical}\:{approach}\:{yields}\: \\ $$$${the}\:{same}\:{result}\:{since}\:\left(\mathrm{1}\right)\:{is}\:{equivalent}\:{to}\:{y}=\mathrm{0}\:{for} \\ $$$${y}=\mathrm{3}^{{log}\left(\mathrm{3}{x}+\mathrm{4}\right)} −\mathrm{4}^{{log}\left(\mathrm{4}{x}+\mathrm{3}\right)} \:{defined}\:{for}\:{x}>−\frac{\mathrm{3}}{\mathrm{4}}. \\ $$$${Numerical}\:{methods}\:{work}\:{as}\:{well}. \\ $$$${Otherwise},\:{I}\:{could}\:{not}\:{find}\:{an}\:{algebraic}\:{means}\: \\ $$$${of}\:{solving}\:{for}\:{x}\:{of}\:{the}\:{form}\: \\ $$$$\frac{{log}\left({ax}+{b}\right)}{{log}\left({cx}+{d}\right)}={e}\:{where}\:{all}\:{of}\:{a},{b},{c},{d}\:{and}\:{e}\:{are} \\ $$$${non}−{zero}\:{reals}. \\ $$

Commented by Rasheed Soomro last updated on 10/Aug/15

I appreciate your approach. But I am doubtful about your claim that there is only one real solution (x=0 ). You correctly derived : ((log(3x+4))/(log(4x+3)))=((log4)/(log3)) Taking few steps more ,we will get, log(3x+4)=((log 4)/(log 3)) .log(4x+3) =log(4x+3)^((log 4)/(log 3)) Taking antilog of both sides 3x+4=(4x+3)^((log 4)/(log 3)) Observe right hand side the index ((log 4)/(log 3)) >0. So for positive 4x+3 theRHS must be real, if I am not wrong. So I think there may be other real solution/s too. I agree with you that there is no algebraic way to solve this equation.But numerical methods may help.

$$ \\ $$$${I}\:{appreciate}\:{your}\:{approach}.\:{But}\:{I}\:{am}\:{doubtful}\:{about}\:{your} \\ $$$${claim}\:{that}\:\:{there}\:{is}\:{only}\:{one}\:{real}\:{solution}\:\left({x}=\mathrm{0}\:\right). \\ $$$$\:\:\:\:{You}\:{correctly}\:{derived}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{log}\left(\mathrm{3}{x}+\mathrm{4}\right)}{{log}\left(\mathrm{4}{x}+\mathrm{3}\right)}=\frac{{log}\mathrm{4}}{{log}\mathrm{3}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:{Taking}\:{few}\:\:{steps}\:{more}\:,{we}\:{will}\:{get}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{log}\left(\mathrm{3}{x}+\mathrm{4}\right)=\frac{{log}\:\mathrm{4}}{{log}\:\mathrm{3}}\:.{log}\left(\mathrm{4}{x}+\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={log}\left(\mathrm{4}{x}+\mathrm{3}\right)^{\frac{{log}\:\mathrm{4}}{{log}\:\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:\:\:{Taking}\:{antilog}\:{of}\:{both}\:{sides} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{x}+\mathrm{4}=\left(\mathrm{4}{x}+\mathrm{3}\right)^{\frac{{log}\:\mathrm{4}}{{log}\:\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:\:{Observe}\:{right}\:{hand}\:{side}\:{the}\:{index}\:\frac{{log}\:\mathrm{4}}{{log}\:\mathrm{3}}\:>\mathrm{0}. \\ $$$$\:\:\:\:\:\:\:\:\:\:{So}\:{for}\:{positive}\:\mathrm{4}{x}+\mathrm{3}\:{theRHS}\:{must}\:{be}\:{real},\:{if}\:{I}\:{am} \\ $$$${not}\:{wrong}.\:{So}\:{I}\:{think}\:{there}\:{may}\:{be}\:{other}\:{real}\:{solution}/{s}\:\:{too}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{I}\:{agree}\:{with}\:{you}\:{that}\:{there}\:{is}\:{no}\:{algebraic}\:{way}\:{to}\:{solve}\: \\ $$$${this}\:{equation}.{But}\:{numerical}\:{methods}\:{may}\:{help}. \\ $$

Commented by Yugi last updated on 25/Jul/15

Considering the form 3x+4=(4x+3)^((log4)/(log3)) ........(∗) ((log4)/(log3))=log_3 4 is an irrational exponent. As such, the following numerical approach to solving for x can be done in order to acquire a polynomial in x (whose real solution should be found to be approximately x=0). Firstly, consider the value of log_3 4. log_3 4=1.2618... So to 2 s.f log_3 4≈1.3=((13)/(10)) ∴ 3x+4=(4x+3)^(13/10) Raising both sides by the power of 10 gives (3x+4)^(10) =(4x+3)^(13) While it is possible to expand those binomials to solve for x in f(x)=0 I think it would be easier to define a function y=(3x+4)^(10) −(4x+3)^(13) (x>−3/4,x∈R) and use numerical methods for estimating its root. Graphing yields an answer the fastest. Now, this gives one estimation for x satisfying (∗). To attain a more accurate estimate, use more accuarte rational approximations to log_3 4, such as ((63)/(50)),((6309)/(5000)), etc. Every result of x based on these approximations have different values but the answers are near zero. What I′ve outlined here is one means of obtaining x without knowing how many real roots exist for (∗). However, in general, if x is a negative real number and y is irrational,then all answers of x^y are complex. Also, if x is positive (x∈R) and y∈Q^� then x^y has only one real answer and all of the rest are complex. Hence, RHS(∗) should yield one real result. (Of course I could be wrong. I did some research but I′m just a student who′s here to learn.)

$${Considering}\:{the}\:{form}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{x}+\mathrm{4}=\left(\mathrm{4}{x}+\mathrm{3}\right)^{\frac{{log}\mathrm{4}}{{log}\mathrm{3}}} \:\:\:\:\:........\left(\ast\right) \\ $$$$\frac{{log}\mathrm{4}}{{log}\mathrm{3}}={log}_{\mathrm{3}} \mathrm{4}\:\:{is}\:{an}\:{irrational}\:{exponent}.\:{As}\:{such},\:{the}\:{following}\:\:{numerical}\:{approach}\:{to} \\ $$$${solving}\:{for}\:{x}\:{can}\:{be}\:{done}\:{in}\:{order}\:{to}\:{acquire}\:{a}\:{polynomial}\:{in}\:{x}\:\left({whose}\:{real}\:{solution}\right. \\ $$$$\left.{should}\:{be}\:{found}\:{to}\:{be}\:{approximately}\:{x}=\mathrm{0}\right).\:{Firstly},\:{consider}\:{the}\:{value}\:{of}\:{log}_{\mathrm{3}} \mathrm{4}. \\ $$$${log}_{\mathrm{3}} \mathrm{4}=\mathrm{1}.\mathrm{2618}...\:{So}\:{to}\:\mathrm{2}\:{s}.{f}\:{log}_{\mathrm{3}} \mathrm{4}\approx\mathrm{1}.\mathrm{3}=\frac{\mathrm{13}}{\mathrm{10}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\:\mathrm{3}{x}+\mathrm{4}=\left(\mathrm{4}{x}+\mathrm{3}\right)^{\mathrm{13}/\mathrm{10}} \: \\ $$$${Raising}\:{both}\:{sides}\:{by}\:{the}\:{power}\:{of}\:\mathrm{10}\:{gives} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{10}} =\left(\mathrm{4}{x}+\mathrm{3}\right)^{\mathrm{13}} \\ $$$${While}\:{it}\:{is}\:{possible}\:{to}\:{expand}\:{those}\:{binomials}\:{to}\:{solve}\:{for}\:{x}\:{in}\:{f}\left({x}\right)=\mathrm{0}\:{I}\:{think}\:{it}\:{would}\:{be}\:{easier}\:{to}\:{define} \\ $$$${a}\:{function}\:{y}=\left(\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{10}} −\left(\mathrm{4}{x}+\mathrm{3}\right)^{\mathrm{13}} \:\:\left({x}>−\mathrm{3}/\mathrm{4},{x}\in\mathbb{R}\right)\:{and}\:{use}\:{numerical}\:{methods} \\ $$$${for}\:{estimating}\:{its}\:{root}.\:{Graphing}\:{yields}\:{an}\:{answer}\:{the}\:{fastest}.\:\: \\ $$$${Now},\:{this}\:{gives}\:{one}\:{estimation}\:{for}\:{x}\:{satisfying}\:\left(\ast\right).\:{To}\:{attain}\:{a}\:{more}\:{accurate} \\ $$$${estimate},\:{use}\:{more}\:{accuarte}\:{rational}\:{approximations}\:{to}\:{log}_{\mathrm{3}} \mathrm{4},\:{such}\:{as}\:\frac{\mathrm{63}}{\mathrm{50}},\frac{\mathrm{6309}}{\mathrm{5000}},\:{etc}. \\ $$$${Every}\:{result}\:{of}\:{x}\:{based}\:{on}\:{these}\:{approximations}\:{have}\:{different}\:{values}\: \\ $$$$\:{but}\:{the}\:{answers}\:{are}\:{near}\:{zero}.\:{What}\:{I}'{ve}\:{outlined}\:{here}\:{is}\:{one}\:{means} \\ $$$${of}\:{obtaining}\:{x}\:{without}\:{knowing}\:{how}\:{many}\:{real}\:{roots}\:{exist}\:{for}\:\left(\ast\right).\:{However}, \\ $$$${in}\:{general},\:{if}\:{x}\:{is}\:{a}\:{negative}\:{real}\:{number}\:{and}\:{y}\:{is}\:{irrational},{then}\:{all}\:{answers}\:{of} \\ $$$${x}^{{y}} \:{are}\:{complex}.\:{Also},\:{if}\:{x}\:{is}\:{positive}\:\left({x}\in\mathbb{R}\right)\:{and}\:{y}\in\bar {\mathbb{Q}}\:{then}\:{x}^{{y}} \:{has}\:{only}\:{one}\:{real} \\ $$$${answer}\:{and}\:{all}\:{of}\:{the}\:{rest}\:{are}\:{complex}.\:{Hence},\:{RHS}\left(\ast\right)\:{should}\:{yield}\:{one}\:{real}\:{result}.\: \\ $$$$\left({Of}\:{course}\:{I}\:{could}\:{be}\:{wrong}.\:{I}\:{did}\:{some}\:{research}\:{but}\:{I}'{m}\:{just}\:{a}\:{student}\:{who}'{s}\:{here}\right. \\ $$$$\left.{to}\:{learn}.\right) \\ $$$$ \\ $$

Commented by Rasheed Soomro last updated on 26/Jul/15

I didn′t say ′′ There are certainly more than one soution ′′. But I said,′′ There may be more than one solution′′ and suggested to confirm using numerical analysis.One other suggestion has come from parkash jain for the confirmation. But during this comment I have drawn the graph.It seems that your claim of ′′ only one real solution is correct ′′. After all we all are learners.

$${I}\:{didn}'{t}\:{say}\:''\:{There}\:{are}\:\boldsymbol{\mathrm{certainly}}\:{more}\:{than}\:{one}\:{soution}\:''. \\ $$$${But}\:{I}\:{said},''\:{There}\:\boldsymbol{\mathrm{may}}\:\boldsymbol{\mathrm{be}}\:{more}\:{than}\:{one}\:{solution}''\:\:{and}\: \\ $$$${suggested}\:{to}\:{confirm}\:{using}\:{numerical}\:{analysis}.{One}\:{other}\: \\ $$$${suggestion}\:{has}\:{come}\:{from}\:{parkash}\:{jain}\:{for}\:{the}\:{confirmation}. \\ $$$$\:\:\:\:\:\:\:\:{But}\:{during}\:{this}\:{comment}\:{I}\:{have}\:{drawn}\:{the}\:{graph}.\boldsymbol{\mathrm{It}}\:\boldsymbol{\mathrm{seems}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{your}}\:\boldsymbol{\mathrm{claim}}\:\boldsymbol{\mathrm{of}}\:\:''\:\boldsymbol{\mathrm{only}} \\ $$$$\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{real}}\:\boldsymbol{\mathrm{solution}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{correct}}\:''.\: \\ $$$${After}\:{all}\:{we}\:{all}\:{are}\:{learners}. \\ $$$$ \\ $$

Commented by Yugi last updated on 26/Jul/15

Yes. The graph showed the existence of one real solution.

$${Yes}.\:{The}\:{graph}\:{showed}\:{the}\:{existence}\:{of}\:{one}\:{real}\:{solution}.\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com