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.

${\mathrm{log}}_{a}x=p$, ${\mathrm{log}}_{b}x=q$ , ${\mathrm{log}}_{abc}x$=r. What is ${\mathrm{log}}_{c}x$?.. It's on my math homework can someone solve it cause I need it.

lydalaszq

Beginner2022-07-09Added 11 answers

So, use the log change of basis formula:

$\mathrm{log}(x)/\mathrm{log}(a)=p\phantom{\rule{0ex}{0ex}}\mathrm{log}(x)/\mathrm{log}(b)=q\phantom{\rule{0ex}{0ex}}\mathrm{log}(x)/\mathrm{log}(abc)=(\frac{\mathrm{log}(a)+\mathrm{log}(b)+\mathrm{log}(c)}{\mathrm{log}(x)}{)}^{-1}=r$

and so $\mathrm{log}(x)/\mathrm{log}(c)=(\frac{\mathrm{log}(c)}{\mathrm{log}(x)}{)}^{-1}=(\frac{\mathrm{log}(a)+\mathrm{log}(b)+\mathrm{log}(c)}{\mathrm{log}(x)}-\frac{\mathrm{log}(b)}{\mathrm{log}(x)}-\frac{\mathrm{log}(a)}{\mathrm{log}(x)}{)}^{-1}=(\frac{1}{r}-\frac{1}{q}-\frac{1}{p}{)}^{-1}$

$\mathrm{log}(x)/\mathrm{log}(a)=p\phantom{\rule{0ex}{0ex}}\mathrm{log}(x)/\mathrm{log}(b)=q\phantom{\rule{0ex}{0ex}}\mathrm{log}(x)/\mathrm{log}(abc)=(\frac{\mathrm{log}(a)+\mathrm{log}(b)+\mathrm{log}(c)}{\mathrm{log}(x)}{)}^{-1}=r$

and so $\mathrm{log}(x)/\mathrm{log}(c)=(\frac{\mathrm{log}(c)}{\mathrm{log}(x)}{)}^{-1}=(\frac{\mathrm{log}(a)+\mathrm{log}(b)+\mathrm{log}(c)}{\mathrm{log}(x)}-\frac{\mathrm{log}(b)}{\mathrm{log}(x)}-\frac{\mathrm{log}(a)}{\mathrm{log}(x)}{)}^{-1}=(\frac{1}{r}-\frac{1}{q}-\frac{1}{p}{)}^{-1}$

pouzdrotf

Beginner2022-07-10Added 4 answers

Using

Change of Base Formula:

${\mathrm{log}}_{a}x=\frac{{\mathrm{log}}_{b}x}{{\mathrm{log}}_{b}a}$

And Product:

${\mathrm{log}}_{a}(xy)={\mathrm{log}}_{a}x+{\mathrm{log}}_{a}y$

And

${\mathrm{log}}_{a}a=1$

You have:

${\mathrm{log}}_{abc}x=\frac{lo{g}_{x}x}{{\mathrm{log}}_{x}abc}=\frac{1}{{\mathrm{log}}_{x}abc}=r$

${\mathrm{log}}_{a}x=\frac{{\mathrm{log}}_{x}x}{{\mathrm{log}}_{x}a}=\frac{1}{{\mathrm{log}}_{x}a}=p\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{x}a=\frac{1}{p}$

${\mathrm{log}}_{b}x=\frac{{\mathrm{log}}_{x}x}{{\mathrm{log}}_{x}b}=\frac{1}{{\mathrm{log}}_{x}b}=q\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{x}b=\frac{1}{q}$

So substituting what you have from above:

$\frac{1}{{\mathrm{log}}_{x}abc}=\frac{1}{{\mathrm{log}}_{x}a+{\mathrm{log}}_{x}b+{\mathrm{log}}_{x}c}=\frac{1}{\frac{1}{p}+\frac{1}{q}+{\mathrm{log}}_{x}c}=r$

$1=r(\frac{1}{p}+\frac{1}{q}+{\mathrm{log}}_{x}c)$

$\frac{1-\frac{r}{p}-\frac{r}{q}}{r}={\mathrm{log}}_{x}c$

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

${(\frac{1}{r}-\frac{1}{p}-\frac{1}{q})}^{-1}$

Change of Base Formula:

${\mathrm{log}}_{a}x=\frac{{\mathrm{log}}_{b}x}{{\mathrm{log}}_{b}a}$

And Product:

${\mathrm{log}}_{a}(xy)={\mathrm{log}}_{a}x+{\mathrm{log}}_{a}y$

And

${\mathrm{log}}_{a}a=1$

You have:

${\mathrm{log}}_{abc}x=\frac{lo{g}_{x}x}{{\mathrm{log}}_{x}abc}=\frac{1}{{\mathrm{log}}_{x}abc}=r$

${\mathrm{log}}_{a}x=\frac{{\mathrm{log}}_{x}x}{{\mathrm{log}}_{x}a}=\frac{1}{{\mathrm{log}}_{x}a}=p\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{x}a=\frac{1}{p}$

${\mathrm{log}}_{b}x=\frac{{\mathrm{log}}_{x}x}{{\mathrm{log}}_{x}b}=\frac{1}{{\mathrm{log}}_{x}b}=q\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{x}b=\frac{1}{q}$

So substituting what you have from above:

$\frac{1}{{\mathrm{log}}_{x}abc}=\frac{1}{{\mathrm{log}}_{x}a+{\mathrm{log}}_{x}b+{\mathrm{log}}_{x}c}=\frac{1}{\frac{1}{p}+\frac{1}{q}+{\mathrm{log}}_{x}c}=r$

$1=r(\frac{1}{p}+\frac{1}{q}+{\mathrm{log}}_{x}c)$

$\frac{1-\frac{r}{p}-\frac{r}{q}}{r}={\mathrm{log}}_{x}c$

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

${(\frac{1}{r}-\frac{1}{p}-\frac{1}{q})}^{-1}$

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