Banguizb

2022-07-08

It's a logarithmic worksheet and O can't solve it.
${\mathrm{log}}_{a}x=p$, ${\mathrm{log}}_{b}x=q$ , ${\mathrm{log}}_{abc}x$=r. What is ${\mathrm{log}}_{c}x$?.. It's on my math homework can someone solve it cause I need it.

lydalaszq

So, use the log change of basis formula:
$\mathrm{log}\left(x\right)/\mathrm{log}\left(a\right)=p\phantom{\rule{0ex}{0ex}}\mathrm{log}\left(x\right)/\mathrm{log}\left(b\right)=q\phantom{\rule{0ex}{0ex}}\mathrm{log}\left(x\right)/\mathrm{log}\left(abc\right)=\left(\frac{\mathrm{log}\left(a\right)+\mathrm{log}\left(b\right)+\mathrm{log}\left(c\right)}{\mathrm{log}\left(x\right)}{\right)}^{-1}=r$
and so $\mathrm{log}\left(x\right)/\mathrm{log}\left(c\right)=\left(\frac{\mathrm{log}\left(c\right)}{\mathrm{log}\left(x\right)}{\right)}^{-1}=\left(\frac{\mathrm{log}\left(a\right)+\mathrm{log}\left(b\right)+\mathrm{log}\left(c\right)}{\mathrm{log}\left(x\right)}-\frac{\mathrm{log}\left(b\right)}{\mathrm{log}\left(x\right)}-\frac{\mathrm{log}\left(a\right)}{\mathrm{log}\left(x\right)}{\right)}^{-1}=\left(\frac{1}{r}-\frac{1}{q}-\frac{1}{p}{\right)}^{-1}$

pouzdrotf

Using
Change of Base Formula:
${\mathrm{log}}_{a}x=\frac{{\mathrm{log}}_{b}x}{{\mathrm{log}}_{b}a}$
And Product:
${\mathrm{log}}_{a}\left(xy\right)={\mathrm{log}}_{a}x+{\mathrm{log}}_{a}y$
And
${\mathrm{log}}_{a}a=1$
You have:
${\mathrm{log}}_{abc}x=\frac{lo{g}_{x}x}{{\mathrm{log}}_{x}abc}=\frac{1}{{\mathrm{log}}_{x}abc}=r$
${\mathrm{log}}_{a}x=\frac{{\mathrm{log}}_{x}x}{{\mathrm{log}}_{x}a}=\frac{1}{{\mathrm{log}}_{x}a}=p\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{x}a=\frac{1}{p}$
${\mathrm{log}}_{b}x=\frac{{\mathrm{log}}_{x}x}{{\mathrm{log}}_{x}b}=\frac{1}{{\mathrm{log}}_{x}b}=q\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{x}b=\frac{1}{q}$
So substituting what you have from above:
$\frac{1}{{\mathrm{log}}_{x}abc}=\frac{1}{{\mathrm{log}}_{x}a+{\mathrm{log}}_{x}b+{\mathrm{log}}_{x}c}=\frac{1}{\frac{1}{p}+\frac{1}{q}+{\mathrm{log}}_{x}c}=r$
$1=r\left(\frac{1}{p}+\frac{1}{q}+{\mathrm{log}}_{x}c\right)$
$\frac{1-\frac{r}{p}-\frac{r}{q}}{r}={\mathrm{log}}_{x}c$
Following from above
${\mathrm{log}}_{c}x=\frac{1}{{\mathrm{log}}_{x}c}=\frac{1}{\frac{1-\frac{r}{p}-\frac{r}{q}}{r}}=\frac{1}{\frac{1}{r}-\frac{1}{p}-\frac{1}{q}}$
Can also be expressed as
${\left(\frac{1}{r}-\frac{1}{p}-\frac{1}{q}\right)}^{-1}$

Do you have a similar question?