Question

PLEASE HELP!! :(
I have this y=mx+B form ->

y=-0.031ln(x) + 0.1306

have to plug in 2 for x.
so it would be
y=-0.031ln(2) + 0.1306
how would I solve this? I forgot the log rules and I need this for a bio lab report. please help me. Thank you so much!

Answer 1

should I find the Ln of 2 first than multiply the -0.031 and add 0.1306?....

Answer 2

y=-0.031ln(x) + 0.1306 is not in the form y = mx+b though

Answer 3

yeah it is.well, technically it won't seem like it but yes. it is.
the m=-0.031
x=2
b=0.1306

Answer 4

alright so if m = -0.031 and b = 0.1306, then

y = mx+b

y = -0.031x+0.1306

Answer 5

assuming it is in y=mx+b form

Answer 6

ln 2=0.693

Answer 7

just plug in this value in place of ln2 and do the rest of operation required and find y

Answer 8

so I find the Ln first than just do the rest?

Answer 9

Well you can always download a graphing utility (graphcalc, or some other scientific calculator) with an (ln) operator and replace y with f(x), so you would be working with a linear function when, x = 2.
f(x) = -0.031ln(x) + 0.1306
f(2) = -0.031ln(2) + 0.1306

And then evaluate when x = 2, at f(2) = ?

Type in the calculator ln(2), multiply by -0.031, then add 0.1306

Answer 10

I was trying to do it with excel but it was weird and didn't give me an answer. so I have to solve it by hand. I just didn't know where the Ln was in regards to the order of operations

Answer 11

Also, considering this is in slope-intercept form, you can graph it relatively simple as well.
Here is the natural log graph, what is y when x = 2?
That is the same question as what is f(2) in the function f(x), when f(x) = -0.031ln(x) + 0.1306

Answer 12

Here is the graph with the cursor just about where x = 2. This is the graphical representation of the solution f(2) to the linear equation shown algebraically as
f(x) = -0.031ln(2) + 0.1306

Answer 13

yeah but I need to find the value for this, another X and divide by 2. that would be slope 1...well for output 1 in this report. thanks for the graph! I actually downloaded that graph software. I plugged in the f(x) equation as you told me, and it's not solving. I'll just do everything in my calc. but what did u get y as ? for the original q after solving everything? I got 0.1090 (after rounding)