Thread: Need help with solving equations with logs!

1. Need help with solving equations with logs!

Hello everyone,

I am a graduate student in a groundwater hydrology course that is requiring that I dust off my math skills. I need help with solving this equation:

y=3log10x

I need to solve so that I can manually plot this on semilog paper and determine the slope.

A friend said that the data would look like this:
y=0, x=1
y=3, x=10
y=6, x=100
y=9, x=1000

Can someone verify this, please?

2. Re: Need help with solving equations with logs!

Pretty much any scientific calculator will have two logarithm buttons, (though they could be on a single button with one of them accessed via a shift key of some sort).
The ln option is for calculating the natural log of the number, that is, the log base e.
The other option, log, is for calculating the log base 10, which is the one that you are after.
So, enter a value for x, calculate its log, and multiply that by 3. That will be the corresponding value for y.
The data that you give is correct, as you may verify. You might though want to fill in some intermediate values !

3. Re: Need help with solving equations with logs!

BobP,

Thanks for the specific instructions on using the calculator as well as the verification!

4. Re: Need help with solving equations with logs!

It seems a little peculiar to choose values of y and then solve for y in order to graph a function. Most of the time, when you are given y as a function of x, you choose values of x and calculate y and that was what Bobp told you how to do. However, you can use the fact that "$\displaystyle log_{10}(x)$" and "$\displaystyle 10^x$" are "inverse functions". That is, if $\displaystyle y= 3log_{10}(x)$, of course, $\displaystyle \frac{y}{3}= log_{10}(x)$ and then, taking 10 to the power of each side, $\displaystyle 10^{y/3}= 10^{log_{10}(x)}= x$.
That's why:
when y= 0, $\displaystyle x= 10^{0/3}= 10^0= 1$
when y= 3, $\displaystyle x= 10^{3/3}= 10^1= 10$
when y= 6, $\displaystyle x= 10^{6/3}= 10^2= 100$
and when y= 9, $\displaystyle x= 10^{9/3}= 10^3= 1000$

and why your friend chose y to be multiples of 3!