Maths Wrong Qs

2Fast2C

I got an assignment and my teacher told me that one of these 2 questions is wrong, as in there is a mistake in it.
I am here sitting hours, trying to figure out which one it is, could some one shed some light,
Thanks

Log25(6x+25)+log5(x+25)=5,XEN

3 to the power of 2x+1 - 13(3 to ther power of x) +14=0

Peregrine

And have you tried and solved both of them?

What did you mean by a mistake in the question? Is there a final answer in the book that is wrong?

Edit: Because I just had a quick look at both of them and they look solvable.

Take Your Pants Off

Nim wrote: »

And have you tried and solved both of them?

What did you mean by a mistake in the question? Is there a final answer in the book that is wrong?

Edit: Because I just had a quick look at both of them and they look solvable.

I am just trying them now aswell

2Fast2C

Well the first one I tried and I got nowhere, second one gives me y squared-13y+14 which is not sovelable without quadratic formula and it gives decimal answers, which you dont usually get for these kind of qs.

Peregrine

I've got the second one done, just trying the first one now.

2Fast2C wrote: »

Well the first one I tried and I got nowhere, second one gives me y squared-13y+14 which is not sovelable without quadratic formula and it gives decimal answers, which you dont usually get for these kind of qs.

You're nearly there with the second one. Check your coefficient for y^2 and try to get the roots again.

Actually, just by looking at the first one, x = 100 is a solution. So the question is, what's wrong with it?

2Fast2C

Nim wrote: »

I've got the second one done, just trying the first one now.



You're nearly there with the second one. Check your coefficient for y^2 and try to get the roots again.

Actually, just by looking at the first one, x = 100 is a solution. So the question is, what's wrong with it?

Well tbh I do not know, we had a test on this and these 2 questions came up, I had a trip that day and I asked for the test qs and the teacher said that she did not count one of the questions that came up, I just could not remember which one she had said, do you mind telling me how you did them both, I cant seem to get past b,
(3^x)^2.3^1 thats as far as I got

Peregrine

2Fast2C wrote: »

Well tbh I do not know, we had a test on this and these 2 questions came up, I had a trip that day and I asked for the test qs and the teacher said that she did not count one of the questions that came up, I just could not remember which one she had said, do you mind telling me how you did them both, I cant seem to get past b,
(3^x)^2.3^1 thats as far as I got

Sure.

I'll use the dot for multiplication btw, not as a decimal point.

[latex]3^{2x+1} - 13.3^x + 14 = 0[/latex]
[latex]3^{2x}.3^1 - 13.3^x + 14 = 0[/latex]
[latex]3.(3^{x})^2 - 13.3^x + 14 = 0[/latex]
That's a quadratic in [latex]3^x[/latex]. I see you substituted [latex]3^x = y[/latex] and solved for y. That's good. When you do that, you get:

[latex]3y^2 - 13y + 14 = 0[/latex]
[latex]y = 2, \frac{7}{3}[/latex]

From there, you solve for x. It's some mad decimal but it works.

As for Q1

[latex]log_{25}(6x+25) +log_5(x+25) = 5[/latex]
[latex]log_{\sqrt{25}}(\sqrt{6x+25}) +log_5(x+25) = 5[/latex]
[latex]log_{5}(\sqrt{6x+25}) +log_5(x+25) = 5[/latex]

Solve from there to get x= 100

You might have to view this post from a computer to read it properly. Or even view it in desktop mode on you phone.

2Fast2C

Nim wrote: »

Sure.

I'll use the dot for multiplication btw, not as a decimal point.

[latex]3^{2x+1} - 13.3^x + 14 = 0[/latex]
[latex]3^{2x}.3^1 - 13.3^x + 14 = 0[/latex]
[latex]3.3^{x^2} - 13.3^x + 14 = 0[/latex]
That's a quadratic in [latex]3^x[/latex]. I see you substituted [latex]3^x = y[/latex] and solved for y. That's good. When you do that, you get:

[latex]3y^2 - 13y + 14 = 0[/latex]
[latex]y = 2, \frac{7}{3}[/latex]

From there, you solve for x. It's some mad decimal but it works.

As for Q1

[latex]log_{25}(6x+25) +log_5(x+25) = 5[/latex]
[latex]log_{\sqrt{25}}(\sqrt{6x+25}) +log_5(x+25) = 5[/latex]
[latex]log_{5}(\sqrt{6x+25}) +log_5(x+25) = 5[/latex]

Solve from there to get x= 100

You might have to view this post from a computer to read it properly. Or even view it in desktop mode on you phone.

Thanks for that, for the log question, when I solve it (squareroot6x +25)(x+25)
I get 2.45xsquared+66.25x+125=0
Is that what you got ?

Peregrine

2Fast2C wrote: »

Thanks for that, for the log question, when I solve it (squareroot6x +25)(x+25)
I get 2.45xsquared+66.25x+125=0
Is that what you got ?

No, it's slightly different. Square both sides and solve for x. This may be the question your teacher was talking about. It does get really messy.

[latex]log_{5}(\sqrt{6x+25}) +log_5(x+25) = 5[/latex] Simplify
[latex]log_{5}(\sqrt{6x+25})(x+25) = 5[/latex]
[latex](\sqrt{6x+25})(x+25) = 5^5[/latex]
[latex](6x+25)(x+25)^2 = 5^{10}[/latex] Squaring both sides
[latex](6x+25)(x^2+50x+625) = 5^{10}[/latex]
[latex]6x^3 + 325x^2 + 5000x - 9750000 = 0[/latex]

This is where it gets tricky. Solving cubic equations is way beyond the Leaving Cert Maths syllabsus. Now, I looked at the initial equation in logs for a bit and realised that x = 100 is a solution. You'll have to do something similar here. Put in numbers til it balances.

[latex]6x^3 + 325x^2 + 5000x = 9750000 [/latex]

You're looking at a pretty big x value there to get 9 million something on the LHS. We know 1 won't work, go to the next order of magnitude, 10. It's still not nowhere near it. Try 100 and it balances. This equation was set up to be easy to solve. I know that's hard to understand and that's probably why your teacher didn't count this one.

Since there are three roots, you might as well do this for good measure:

Divide [latex]6x^3 + 325x^2 + 5000x - 9750000 = 0[/latex] by [latex]x-100[/latex] which is a factor since [latex]x=100[/latex] is a root. You get a quadratic.

So, [latex](x-100)(6x^2 +925x +97500) = 0[/latex]
That quadratic has no real roots and since x ∈ N, x = 100 is the only solution.

Phew. I hope that wasn't too confusing.

[latex][/latex]

2Fast2C

Nim wrote: »

No, it's slightly different. Square both sides and solve for x. This may be the question your teacher was talking about. It does get really messy.

[latex]log_{5}(\sqrt{6x+25}) +log_5(x+25) = 5[/latex] Simplify
[latex]log_{5}(\sqrt{6x+25})(x+25) = 5[/latex]
[latex](\sqrt{6x+25})(x+25) = 5^5[/latex]
[latex](6x+25)(x+25)^2 = 5^{10}[/latex] Squaring both sides
[latex](6x+25)(x^2+50x+625) = 5^{10}[/latex]
[latex]6x^3 + 325x^2 + 5000x - 9750000 = 0[/latex]

This is where it gets tricky. Solving cubic equations is way beyond the Leaving Cert Maths syllabsus. Now, I looked at the initial equation in logs for a bit and realised that x = 100 is a solution. You'll have to do something similar here. Put in numbers til it balances.

[latex]6x^3 + 325x^2 + 5000x = 9750000 [/latex]

You're looking at a pretty big x value there to get 9 million something on the LHS. We know 1 won't work, go to the next order or magnitude, 10. It's still not nowhere near it. Try 100 and it balances. This equation was set up to be easy to solve. I know that's hard to understand and that's probably why your teacher didn't count this one.

Since there are three roots, you might as well do this for good measure:

Divide [latex]6x^3 + 325x^2 + 5000x - 9750000 = 0[/latex] by [latex]x-100[/latex] which is a factor since [latex]x=100[/latex] is a root. You get a quadratic.

So, [latex](x-100)(6x^2 +925x +97500) = 0[/latex]
That quadratic has no real roots and since x ∈ N, x = 100 is the only solution.

Phew. I hope that wasn't too confusing.

[latex][/latex]

Thank you so much for taking the time out.
I understand this perfectly now.
My gratitude towards you is beyond the word thanks.

Peregrine

2Fast2C wrote: »

Thank you so much for taking the time out.
I understand this perfectly now.
My gratitude towards you is beyond the word thanks.

You're welcome