logarithms
I understand how to solve a problem such as $6^{2x+3}=11$ by using natural log, but the question thats tripping me up is $3^{1-x}=7^x$. Mathway and Wolfram Alpha tells me what the answer is, but I cant seem to reproduce that answer. Any help?
The Question im being asked is : Solve the exponential equation. Write the exact answer with natural logarithms and then approximate the result correct to three decimal places.
$$3^{1-x}=7^x$$
$$\log {m}=\log {n}+\frac{3}{2} \log {\left(1+\frac{v}{m^2}\right)}$$ $$\log {m}=\log {n}+ \log {\left(\left(1+\frac{v}{m^2}\right)^{\frac{3}{2}}\right)}$$ $$\log {m}=\log {\left({n\cdot\left(1+\frac{v}{m^2}\right)^{\frac{3}{2}}}\right)}$$ Exponentiate both sides: $$m=n\cdot\left(1+\frac{v}{m^2}\right)^{\frac{3}{2}}$$ $$m^2=n^2\cdot\left(1+\frac{v}{m^2}\right)^3$$ $$m^2=n^2\cdot\left(1+\frac{3v}{m^2}+\frac{3v^2}{m^4}+\frac{v^3}{m^6}\right)$$ $$m^2=\frac{n^2}{m^6}\left(m^6+3v\cdot m^4+3v^2 \cdot m^2+v^3\right)$$ $$m^8=n^2 m^6+3vn^2 m^4+ 3n^2v^2 m^2+n^2v^3$$ Substitute $m^2=u$ and you will obtain a quartic expression. $$u^4-n^2u^3-3vn^2 u^2-3n^2 v^2 u-n^2v^3=0$$ This is going to be a long solution, however you can use the general formula for the solution to a quartic equation.
Wikipedia suggests using substitutions in order to solve it.
The full formula without substitution for $ax^4+bx^3+cx^2+dx+e$ is below (I cannot write it in $\LaTeX$ because it is so long).
This suggests that your solution in Wolfram Alpha is probably correct.
The expression in question is
$$\frac{4\ln^3(11)}{11}\;.$$
In this context $\ln^3(11)$ means $(\ln 11)^3$, just as $\sin^2\theta$ normally means $(\sin\theta)^2$; this is approximately $2.3978953^3$, or about $13.787662$. Now multiply that by $4$ and divide by $11$ to get about $5.0136953$.
Best Answer
Taking natural log from both side gives: $$\ln{3}(1-x) = x\ln{7}\\ x(\ln{3} + \ln{7}) = \ln{3}\\ x = \frac{\ln{3}}{\ln{3} + \ln{7}}=\frac{\ln{3}}{\ln21}$$ Then use a calculator to get your desired answer as 3 decimal places.