How do you solve a problem that has a divided variable as a base of a logarithm? My problem is Loga(x)/Log a/b(x). I guess that would be read; Log 'a' of x divided by Log 'a/b' of x. Thanks
In loga(x), which one is base? a or x? If a is the base, use change of base rule. You can take any base. So the problem becomes
Loga(x)/Log a/b(x) = [log(x)/log(a)]/[log(x)/log(a/b)] = log(a/b)/log(a) = [log(a) - log(b)]/log(a) = 1 - log(b)/log(a) = 1 - loga(b) i.e 1 - log b to the base a.
Dear ThanksForHelp,
Let, $\displaystyle \log_{\frac{a}{b}}x=y$
$\displaystyle \left(\frac{a}{b}\right)^y=x$
$\displaystyle a^y=b^{y}x$
$\displaystyle y=y\log_{a}b+\log_{a}x$
$\displaystyle y=\frac{\log_{a}x}{1-\log_{a}b}$
Therefore, $\displaystyle \log_{\frac{a}{b}}x=\frac{\log_{a}x}{1-\log_{a}b}$
I hope this will help you.
Hello, ThanksForHelp!
Are you familiar with the Base-Change Formula?
. . $\displaystyle \log_b(x) \:=\:\frac{\ln x}{\ln b}$
$\displaystyle \text{Simplify: }\:Y \;=\; \frac{\log_a(x)}{\log_{\frac{a}{b}}(x)}$
We have: .$\displaystyle \begin{Bmatrix}\log_a(x) &=& \dfrac{\ln x}{\ln a} \\ \\[-3mm] \log_{\frac{a}{b}}(x) &=& \dfrac{\ln x}{\ln\frac{a}{b}} \end{Bmatrix}$
$\displaystyle \text{Then: }\;Y \;=\;\frac{\,\dfrac{\ln x}{\ln a}\,}{\dfrac{\ln x}{\ln\frac{a}{b}}} \;=\;\frac{\ln x}{\ln a}\cdot\frac{\ln\frac{a}{b}}{\ln x}} \;=\;\frac{\ln\frac{a}{b}}{\ln a}$
$\displaystyle \text{Therefore: }\;Y \;=\;\frac{\ln a - \ln b}{\ln a}\;\text{ or }\;1 - \frac{\ln b}{\ln a}$