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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.

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    $\begingroup$ I'd recommend rewriting each equation of the form $\log_rs=t$ as $r^t=s$, and then using what you know about laws of exponents (and using $84=12\times7$). $\endgroup$ – Gerry Myerson May 28 '19 at 0:19
  • $\begingroup$ I've tried to do it but the farthest I got was 7/a and (b*log base 5 of 11)/a $\endgroup$ – Matthew Zaldaña May 28 '19 at 0:22
  • $\begingroup$ I don't know what you mean when you write that you "got" $7/a$. Dealing with equations, you should get equations, not expressions. I have no idea where you got an 11, and my suggestion was, and is, to get rid of all logarithms, replacing them with exponentials. Try again! but explain what you're trying. $\endgroup$ – Gerry Myerson May 28 '19 at 0:24
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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:

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

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  • $\begingroup$ Wow thanks, I completely forgot that you can factor. $\endgroup$ – Matthew Zaldaña May 28 '19 at 0:29
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$84=7*12$, so $\log_5 {84}=\log_5 {7}+\log_5 {12}$. Now, $\log_5 {12}=\frac{1}{\log_{12}5}$ and $\log_5 {7}=\frac{\log_{12}7}{\log_{12}5}$.

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

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

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

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

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    $\begingroup$ This is a full answer, not a hint... $\endgroup$ – lhf May 28 '19 at 0:26
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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 $w=12$ $$\log_{5}(12)+\log_{5}(7)=\frac{\log_{12}(12)}{\log_{12}(5)}+\frac{\log_{12}(7)}{\log_{12}(5)}$$

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