Find log base 5 of 84 when other logs are given

Question

Been trying to help out a student with a logarithms problem that has me stumped for a week now. I know what the answer is but I don't know how to get to it. It goes like this:

Suppose $\log_{12}5 =a$ and $\log_{12}7=b$.

I am supposed to use rules and properties of logarithms to write this in terms of a and b: $\log_{5}84$.

I know that the answer is: $(1 + b)/a$ but I can't seem to get to it.

Answer 1

There are four important things to remember here.

The first is that you can factor numbers and here we have $84 = 12\cdot 7$

The second is that $\log_x(y\cdot z) = \log_x(y) + \log_x(z)$

The third is that $\log_x(y) = \dfrac{\log_z(y)}{\log_z(x)}$

Finally, remember that $\log_x(x)=1$

These are true for all positive real values of $x,y,z$ different than $1$.

So... we have $\log_{12}(5)=a$ and $\log_{12}(7)=b$

Using these, since these logarithms are the same base we can find $\log_5(7)$ as being $\frac{b}{a}$, but that isn't quite what we are interested in finding, but it is close.

Rather, we note that $\log_{12}(84)=\log_{12}(12\cdot 7) = \log_{12}(12)+\log_{12}(7)=1+b$

Now we can do our division to change the logarithm's base to $5$ and we get:

$$\log_5(84) = \dfrac{\log_{12}(84)}{\log_{12}(5)} = \dfrac{1+b}{a}$$

Answer 2

$$\ln(84)=\ln(12\times 7)=\ln(12)+\ln(7)$$

$$\log_{5}(84)=\frac{\ln(12)}{\ln(5)}+\frac{\ln(7)}{\ln(5)}$$

$$=\frac 1a+\frac{\ln(7)}{\ln(12)}\frac{\ln(12)}{\ln(5)}$$

$$=\frac 1a+\frac ba$$

Answer 3

$84=7*12$, so $\log_5 {84}=\log_5 {7}+\log_5 {12}$. Now, $\log_5 {12}=\frac{1}{\log_{12}5}$ and $\log_5 {7}=\frac{\log_{12}7}{\log_{12}5}$.

Answer 4

Note that $$\log_{5}84=\log_{5}(12\cdot7)=\log_{5}(12)+\log_{5}(7)$$ Now use that $$\log_{x}y=\frac{\log_{w}y}{\log_{w}x}$$ then if $w=12$ $$\log_{5}(12)+\log_{5}(7)=\frac{\log_{12}(12)}{\log_{12}(5)}+\frac{\log_{12}(7)}{\log_{12}(5)}$$