y= bc^x goes through (2, 16) and (3, 128)

I tried to substitute y with 1 and x with 0, since the graph crosses (0,1), and I got b = 1 and c = 4 as my answers. The textbook says the answers are actually b = 0.25, and c = 8. How would I go about getting this and using both the coordinates in the question?

I tried to substitute y with 1 and x with 0, since the graph crosses (0,1), and I got b = 1 and c = 4 as my answers. The textbook says the answers are actually b = 0.25, and c = 8. How would I go about getting this and using both the coordinates in the question?

(edited 1 year ago)

Original post by Bleepbloopbop

y= bc^x goes through (2, 16) and (3, 128)

I tried to substitute y with 1 and x with 0, since the graph crosses (0,1), and I got b = 1 as my answer. The textbook says the answers are actually b = 0.25, and c = 8. How would I go about getting this and using the coordinates in the question?

I tried to substitute y with 1 and x with 0, since the graph crosses (0,1), and I got b = 1 as my answer. The textbook says the answers are actually b = 0.25, and c = 8. How would I go about getting this and using the coordinates in the question?

Ah, the problem you had is that the graph doesn't actually passes through (0,1). It would be true for the case $y=a^x$, but it generally doesn't hold if you add a multiplier to the exponential function. In fact if you plug in x=0 into $y=bc^x$, all you can infer is y=b in this case, which based on the given, b could really be anything.

Now to solve the question, all you need to know is "what does it mean for a graph, with a given equation, to pass through some point (a,b)?" Then it's all algebra. Feel free to ask away if you need further help!

(edited 1 year ago)

Original post by Bleepbloopbop

y= bc^x goes through (2, 16) and (3, 128)

I tried to substitute y with 1 and x with 0, since the graph crosses (0,1), and I got b = 1 and c = 4 as my answers. The textbook says the answers are actually b = 0.25, and c = 8. How would I go about getting this and using both the coordinates in the question?

I tried to substitute y with 1 and x with 0, since the graph crosses (0,1), and I got b = 1 and c = 4 as my answers. The textbook says the answers are actually b = 0.25, and c = 8. How would I go about getting this and using both the coordinates in the question?

Use the two points you are given - the graph doesn't go through (0,1)

Original post by tonyiptony

Ah, the problem you had is that the graph doesn't actually passes through (0,1). It would be true for the case $y=a^x$, but it generally doesn't hold if you add a multiplier to the exponential function. In fact if you plug in x=0 into $y=bc^x$, all you can infer is y=b in this case, which based on the given, b could really be anything.

Now to solve the question, all you need to know is "what does it mean for a graph, with a given equation, to pass through some point (a,b)?" Then it's all algebra. Feel free to ask away if you need further help!

Now to solve the question, all you need to know is "what does it mean for a graph, with a given equation, to pass through some point (a,b)?" Then it's all algebra. Feel free to ask away if you need further help!

Thanks for the help, I'm still struggling to understand. I tried to solve for bc, which gave me 16 = bc^2 = sqrt16 = sqrt b x c.

I tried to rearrange that to make sqrt 16 / sqrt b = c, and substitute it into y = b x (4/ sqrt b)^x, I tried it with c as well, but I am continuing to get the wrong answers.

(edited 1 year ago)

Original post by Relayer

Only the c is squared (bc are not in brackets).

Oh thanks, I wasn't too sure so I tried both ways and even then I end up with 4/b = c and can't figure it out.

Original post by Bleepbloopbop

Thanks for the help, I'm still struggling to understand. I tried to solve for bc, which gave me 16 = bc^2 = sqrt16 = sqrt b x c.

I tried to rearrange that to make sqrt 16 / sqrt b = c, and substitute it into y = b x (4/ sqrt b)^x, I tried it with c as well, but I am continuing to get the wrong answers.

I tried to rearrange that to make sqrt 16 / sqrt b = c, and substitute it into y = b x (4/ sqrt b)^x, I tried it with c as well, but I am continuing to get the wrong answers.

Well, it seems like you understood that your assumption was wrong, so that's progress! Good job.

While so far your steps are mathematically correct, I think you got a bit confused with the task at hand though. I'm not even sure how you end up with "answers", let alone "wrong ansswers".

So ask yourself this: What exactly are you trying to find? Is it x,y? or is it b,c?

Also, very rarely does the question give you information that are of irrelevance. Please look at the question again and see what haven't you used yet.

Original post by tonyiptony

Well, it seems like you understood that your assumption was wrong, so that's progress! Good job.

While so far your steps are mathematically correct, I think you got a bit confused with the task at hand though. I'm not even sure how you end up with "answers", let alone "wrong ansswers".

So ask yourself this: What exactly are you trying to find? Is it x,y? or is it b,c?

Also, very rarely does the question give you information that are of irrelevance. Please look at the question again and see what haven't you used yet.

While so far your steps are mathematically correct, I think you got a bit confused with the task at hand though. I'm not even sure how you end up with "answers", let alone "wrong ansswers".

So ask yourself this: What exactly are you trying to find? Is it x,y? or is it b,c?

Also, very rarely does the question give you information that are of irrelevance. Please look at the question again and see what haven't you used yet.

I finally got the correct answers by solving as simultaneous equations. I shunned this at the start because I thought I could only do that when I add/subtract 2 unknowns rather than multiply. Thank you for the help once again.

Original post by Bleepbloopbop

I finally got the correct answers by solving as simultaneous equations. I shunned this at the start because I thought I could only do that when I add/subtract 2 unknowns rather than multiply. Thank you for the help once again.

Well done! If multiplying(/dividing) 2 equations to eliminate variables doesn't feel right, you can always go back to substitution instead, which works just as well (the problem you had before was simply not using the given information fully).

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