Challenge - Combining into a single logarithm

[chiurox]

So, my precalc teacher gave us this question just for fun at the end of class and I always get stuck in the last steps.

1/2ln(x^2 + 1) - 4ln(1/2) - 1/2[ln(x-4) + lnx]

Can anyone explain me how to get the answer?

 

[galactus]

Remember the rules of logs:

\(\displaystyle \L\\ln(a)-ln(b)=ln(\frac{a}{b})\)

\(\displaystyle \L\\ln(a)+ln(b)=ln(ab)\)

\(\displaystyle \L\\bln(a)=ln(a^{b})\)

 

[soroban]

Hello, chiurox!

Galactus gave you the three rules: \(\displaystyle \:\begin{array}{cccc}\log(A)\,+\log(B) & = & \log(A\cdot B) & \;[1] \\ \log(A)\,-\,\log(B) & = & \log\left(\frac{A}{B}\right) &\;[2]\\ n\cdot\log(A) & = & \log\left(A^n\right) & \;[3] \end{array}\)


We have: \(\displaystyle \:\frac{1}{2}\cdot\ln(x^2\,+\,1)\,-\,4\cdot\ln\left(\frac{1}{2}\right)\,-\,\frac{1}{2}\underbrace{\left[\ln(x\,-\,4)\,+\,\ln(x)\right]}\)

Using [1]: \(\displaystyle \:\frac{1}{2}\cdot\ln(x^2\,+\,1)\,-\,4\cdot\ln\left(\frac{1}{2}\right)\,-\,\frac{1}{2}\cdot\ln\left[x(x\,-\,4)\right]\)


Using [3]: \(\displaystyle \:\ln(x^2\,+\,1)^{\frac{1}{2}}\,-\,\ln\left(\frac{1}{2}\right)^4\,-\,\ln\left[x(x\,-\,4)\right]^{\frac{1}{2}}\)

. . . . . . \(\displaystyle =\;\underbrace{\ln(x^2\,+\,1)^{\frac{1}{2}}\,-\,\ln\left(\frac{1}{16}\right)}\,-\,\ln\left[x(x\,-\,4)\right]^{\frac{1}{2}}\)

Using [2]: \(\displaystyle \;\;\ln\left[\frac{(x^2\,+\,1)^{\frac{1}{2}}} {\frac{1}{16}}\right] \,-\,\ln\left[x(x\,-\,4)\right]^{\frac{1}{2}}\)

. . . . . . . \(\displaystyle =\;\ln\left[16\left(x^2\,+\,1\right)^{\frac{1}{2}}\right]\,-\,\ln\left[x(x\,-\,4)\right]^{\frac{1}{2}}\)

Using [2]: \(\displaystyle \;\;\L\ln\left(\frac{16(x^2+1)^{\frac{1}{2}}}{\left[x(x\,-\,4)\right]^{\frac{1}{2}}}\right)\)

 

[chiurox]

Thanks a lot! Now that I see the whole process, I no longer have any problems with those.