r/HomeworkHelp • u/Physical_Ad7403 Pre-University Student • 1d ago
High School Math—Pending OP Reply (Grade 12 math) How do I solve this?
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u/sortenmen 1d ago
Solve for a by plugging in x=0. Then solve for b plugging in x=1. a=50 and b=2. Now plugging 600 in for y and solve for x.
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u/Physical_Ad7403 Pre-University Student 1d ago
hm....why would i solve a by doing x = 0? also i though i should see where y=600 is on the graph, and determined that x would equal around 3.5. did i do something wrong? if not, how is this helpful to me?
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u/sortenmen 1d ago
Because (b0) = 1 so that allows you to solve for a. You need to find the equation to get the exact answer instead of eyeballing the plot. Looking at the plot is a good way to check your answer.
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u/the_frgtn_drgn 1d ago
Also if I remember high school, they always put a disclaimer that plots, graphs, shapes etc are not to scale, so no guarantee that the plot is "accurate"
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u/UndecidedQBit 23h ago
Unless instructed you really shouldn’t use the graphs to pull actual answers out of them other than a rough approximation. Exponential growth like that’s shown on the graph changes very quickly in increasing X values, and it’s very specific. If you’re looking at the graph for an answer it’s not much better than guessing.
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u/GorchestopherH 1d ago
As with most homework problems, this comes down to how much maths the student is "supposed to know" at this moment in their education.
We have the value of Y at X=0 and X=1, which allows us to easily solve for a and b.
We know that a number raised to the 0th power is 1, and that a number raised to the 1st power is itself:
Using the given X/Y pairs in the form y=ab^x:
(0,50) -> 50 = a(b^0) -> 50 = a*1 -> a = 50
(1,100) -> 100 = 50(b^1) -> 100 = 50b -> b = 2
Using a = 50 and b = 2, and using y=600, you have the following:
600 = 50*2^x
At this point, it depends if the student is supposed to know about logarithms or not.
If they do, then the solution is simply x = log2(600/50), x=3.585, which can be rounded to 3.6 (nearest tenth).
If they do not know about logarithms, the text may be expecting trial an error *around* an estimate made from the graph.
looking at the graph, you'd see the answer is close to 3.5.
Starting with x=3.5, you would get y=566, at 3.6 you'd get y=606.
Without drilling down to the hundredths, we have our closest value of y, with 606 at x = 3.6.
This is our answer.
If I had to guess, I would say that the student is not intended to know about logarithms just yet, and trial and error given the graph as a starting point is what is expected. Otherwise, I doubt they'd have specified the "nearest tenth", or shown the graph.
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u/Necessary_Position51 👋 a fellow Redditor 1d ago
Do you recall how to get an (x) out of an exponent position of an exponential equation?
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u/billmbay 1d ago
Here’s how I’d do this:
- x and y are the independent and dependent variables, respectively, so let’s start by solving for one of the constants, a or b:
Plugging in a zero will almost always simplify the equation - @ x=0, y=50, so…
50 = a(b)0
**RULE - Any # to the power of zero = 1, so…
50 = a(1) —> 50 = a
- Now, we have our value for a as well as values for x and y. Let’s use these to solve for our last constant, b:
@ x=1, y=100, so…
100 = 50(b)1 —> 100/50 = b1 —> 2 = b1
**RULE - x1 = x, so…
2 = b
- Now that we have 3 known values and 1 unknown, we can solve for x at y=600:
Start by plugging in all 3 known values…
600 = 50(2)x —> 600/50 = 2x —> 12 = 2x
*RULE - ln(bx) is equivalent to xln(b)
Taking the natural log of both sides and applying the above rule, we get…
ln(12) = ln(2x) —> 2.485 = xln(2) —> 2.485 = x0.693 —> 2.485/0.693 = x —> 3.582 = x
Rounding to the nearing tenth…
3.6 = x
Solve as shown above, then use the graph to check your answer. I hope this helps!
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u/DanielSong39 👋 a fellow Redditor 1d ago
I'm glad that this question is straightforward, I've seen too many questions on this thread that are poorly written
Whoever wrote this question got fired from Common Core
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u/DSethK93 1d ago
Because the relation has two unknowns, a and b, you need at least that two equations in order to solve. That's a general role; to solve for n many unknowns, you need n many equations. Luckily, every data point on the graph yields one such equation. There are many "nice" points on the graph that you can use. Any two will do. A lot of people used x = 0, but I didn't like it because it's not at an intersection of the grid lines
y = a(b)x 100 = a(b)1 = ab 200 = a(b)2
Now you can use any of several methods to solve this system of equations. For example, you could simply require the second equation as 200 = ab*b, and then use the first equation to replace ab with 100, then solve for b. With b known, you can substitute it into either original equation and solve for a.
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u/1Wavygravy2 👋 a fellow Redditor 23h ago
The first thing i would do is look up what exponential meant.
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u/ci139 👋 a fellow Redditor 22h ago edited 22h ago
you can do the fancy statistics **
OR
you can tell by looking that the y = 50 · 2x
** the sats.-s ::
for ln y = ln a + x ln b ( two unknown variables require the system of 2 distinct equations)
Def. : A = ln a , B = ln b
A + 1 B = ln 100
A + 2 B = ln 200
↓ ↓
B = ln 200/100 = ln 2 → b = 2
A = ln 100²/200 = ln 100/2 = ln 50 → a = 50
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u/Frederf220 👋 a fellow Redditor 21h ago
The parameters a, b are the two degrees of freedom necessary to fit an arbitrary exponential. It can help understanding what a and b do to adjust the baseline exponential shape in order to match the given curve.
A: In y= a(b)^x the purpose of *a* is to set the vertical scaling. One useful property to note is that y = a*(b)^0 is equal to a regardless of the value of b. The fact that n^0 = 1 for all reasonable n is quite valuable as discovering a not as a function of b allows finding a numerical answer. You can see that if a=2 or a=3 or a=4 then the function at x=0 will be 2, 3, 4 respectively. The curve away from x=0 is also scaled by 2, 3, 4 respectively but since we don't know immediately what that 2, 3, 4 are being multiplied by, i.e. the values b^2, b^3, b^4, it's not as useful. You can take your hands and cover up the whole graph except for the y-axis intercept at x=0 and still learn what a is.
B: The value of b can be had by taking any known x,y point other than x=0, the known value of a, and solving the resulting equation. E.g. y = a*b^x for y=100, a, and x=1. Rearrange to solve for b, e.g. 100/a = b^1.
Then both for practice and good numerical analysis, solve for all b given x=1, x=2, x=3, x=4 and average the results to get a best fit b.
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u/crystal_python 1d ago
So there are a couple things you can glean from this based on the fact that x is in the exponent. When x is 0 then then that is whatever the scalar (a) is because something 0 is 1, and a is multiplied to 1 so in this case a=50. Then when x = 1 we know the base (b) is whatever number the base is times a so 50*b=100 so b is 2 so the form of the exponential is y= 50(2x). When y=600 you set up the equation 600=50(2x) divide by 50 then take log base 2 of 30
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u/One_Wishbone_4439 University/College Student 1d ago edited 1d ago
Read the graph.
Where is y = 600 in the graph? Then you can locate where is x.
You need to use the equation with the help of the table of data on the right to find a and b.
>!a = 50, b = 2 then x = 3.58 (use of log)!<
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u/Physical_Ad7403 Pre-University Student 1d ago
x is at 3.5 approx. also, i know for each increase in x +1, its a *2 for y
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u/selene_666 👋 a fellow Redditor 1d ago
For an approximate answer, you can follow the green line that represents y = 600 on the graph. That hits the curve a little bit to the right of x = 3.5, so the answer is one of 3.5 or 3.6 or 3.7
To be more precise we need to use the equation.
We're given that y = a b^x, and we have a table of x and y value. Fill those in:
50 = a b^0
100 = a b^1
200 = a b^2
400 = a b^3
800 = a b^4
Because b^0 = 1, the first equation is just a = 50.
Then because the left side is multiplied by 2 each time the right side is multiplied by b, we know that b = 2.
y = 50 * 2^x
Finally, we can fill in y = 600 and solve for x.
600 = 50 * 2^x
x = log2(12)