help with algebra 2!

[liveandlove435]

help with factoring/solving log equations!

factor: 8x^2(x-6)^3 - 4x(x-6)^4


solve: log(x^2)9 - log(x)36 = 1


factor :
x^3 - (2+i)x^2 + 2ix
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x^3 - (2-i)x^2 - 2ix


I cannot figure out how to do these problems. any help would be appreciated!!

 

[tkhunny]

The one with the logarithms is incomprehensible. Please try again and show your best efforts.

For the first, find common factors in the two terms.

8 compared to 4 ==> 4
x compared to x^2 ==> x
etc.

Show us what you get.

 

[soroban]

Hello, liveandlove435!

\(\displaystyle \text{Solve: }\:\log_{x^2}9 - \log_x36 \:=\:1\)

Note that \(\displaystyle x\) must be positive and \(\displaystyle \ne 1.\)


Use the Base-Change formula.


We have: .\(\displaystyle \dfrac{\log 9}{\log x^2} - \dfrac{\log36}{\log x} \:=\:1 \quad\Rightarrow\quad \dfrac{\log 9}{2\log x} - \dfrac{\log36}{\log x} \:=\:1\)


Multiply by \(\displaystyle 2\log x:\;\;\log 9 - 2\log36 \:=\:2\log x\)


We have: .\(\displaystyle \log x^2 \:=\:\log 9 - \log36^2 \:=\:\log\left(\frac{9}{36^2}\right)\)

. . . . . . . .\(\displaystyle \log x^2 \:=\:\log\left(\frac{1}{144}\right) \)

. . . . . . . . . ..\(\displaystyle x^2 \:=\:\frac{1}{144}\)

. . . . . . . . . . .\(\displaystyle x \:=\:\pm\frac{1}{12}\)


We must discard the negative root.

Therefore: .\(\displaystyle x \,=\,\dfrac{1}{12}\)