SOLUTION: Evaluate the logarithmic equation for three values of x that are less than -1, three values of x that are between 0 and -1, and at x = -1. Show your work. Use the resulting ordered

Question 285176: Evaluate the logarithmic equation for three values of x that are less than -1, three values of x that are between 0 and -1, and at x = -1. Show your work. Use the resulting ordered pairs to plot the graph; State the equation of the line asymptotic to the graph (if any).
y = -log_3.5 (-x)
First, I rewrote the problem to this:
-3.5y = -x
Then substituted values in for 'x'.
Is this correct or is there a much simplier way? Then how do I solve the asymptote?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Evaluate the logarithmic equation for three values of x that are less than -1, three values of x that are between 0 and -1, and at x = -1. Show your work.
--------------------
y = -log_3.5 (-x)
------
I put the equation in my calculator as -[log(-x)/log(3.5)]
I read the y=values off the "Table" on the computer:
I got:
(-2,-0.5533)
(-3,-0.877)
(-4,-1.107)
(-1,0)
(0,error)
(1,error)
(2,error)
-------------
These "error" responses result because there is no
log value for negative numbers.
=======================================
=======================================
y = -log_3.5 (-x)
Writing the equation in exponential form you would get:
-y = log(base3.5)(-x)
3.5^(-y) = -x
----------------------
The asymptote is x = 0 (the y-axis).
===========================================
Cheers,
Stan H.
===========================================

Use the resulting ordered pairs to plot the graph; State the equation of the line asymptotic to the graph (if any).
y = -log_3.5 (-x)
First, I rewrote the problem to this:
-3.5^y = -x
Then substituted values in for 'x'.
Is this correct or is there a much simplier way? Then how do I solve the asymptote?

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