# Thread: How would I graph this Logarithmic function?

1. ## How would I graph this Logarithmic function?

$\displaystyle \log_{8}(x) + 4$

It's just tricky to input it in my calculator

2. Your calculator most likely has the button 'log' implying base 10 or 'ln' implying base e. Here is a property that will help you graph logarithmic functions with different bases:

$\displaystyle \log_{a} (b) = \frac{\log_{c} b}{\log_{c}a}$ where c can be any number.

Since your calculator probably deals with base 10 and e, those are the values you will use for c, i.e.:

$\displaystyle \log_{8} (x) + 4 \: \: = \: \: \frac{\log(x)}{\log(8)} + 4 \: \: = \: \: \frac{\ln(x)}{\ln(8)} + 4$

3. Originally Posted by mankvill
$\displaystyle \log_{8}(x) + 4$

It's just tricky to input it in my calculator
if drawing by hand is what you have to do, here's what you need to know

ALL log graphs of the form $\displaystyle y = \log_a x$ (a > 1) look like the one below. they all pass through the point (1,0), and are asymptotic to the y-axis, and have domain $\displaystyle (0, \infty)$.

with your graph, you are shifting the log up by 4 units (standard transformation rule when you add a positive constant).

now, this shift may affect the intercepts. in this case, it affects the x-intercept only. no worries, you can calculate the new intercept.

for the x-intercept, set y = 0 in your function and solve for x. then just draw a curve that looks like the one below, but instead of passing through (1,0), it passes through the new x-intercept you found.

4. Originally Posted by Jhevon
if drawing by hand is what you have to do, here's what you need to know

ALL log graphs of the form $\displaystyle y = \log_a x$ look like the one below.

...
I don't want to pick at you but the graphs are only similar to your's if a > 1.

5. Originally Posted by earboth
I don't want to pick at you but the graphs are only similar to your's if a > 1.
i assumed that went without saying. i always assume a > 1, you don't see logs with bases less than 1 everyday. i will mention that though. thanks for looking out.