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Stuck on maths questions

Can anyone help me on these two questions, I can't seem to get an answer.

Thanks!
What you have done so far in terms of working?
you for b (the solve one) try take logs of both sides with base 3. Then rearrange to get x,

Hint for A - substitute the value of A into A^2

Hint for B - (AB)^2 - the squared effects both terms

Hints in general
When the squared is outside the bracket so say (Ax)^2, the squared will affect the A and the x leaving you with A^2x^2,

Hope this helps! I'll be happy to help if you have any further questions
Original post by iloading22
Can anyone help me on these two questions, I can't seem to get an answer.

Thanks!


Part b, multiply both sides by lnln or loglog. Use the law ln(a)bbln(a)ln(a)^{b} \equiv b \cdot ln(a) to help you simplify and rearrange.

Part a, if A=15x2A=\frac{1}{5}x^{2} then A2=(15x2)2A^{2} = (\frac{1}{5}x^{2})^{2}. Use the rule (ab)nanbn(ab)^{n} \equiv a^{n}b^{n}

Part b, (AB)2(AB)2(15x25x12)2(AB)^{2} \equiv (A \cdot B)^{2} \Rightarrow (\frac{1}{5}x^{2} \cdot \sqrt5 x^{-\frac{1}{2}})^{2}. I've just substituted in A and B, that's all because you know that A=15x2A=\frac{1}{5}x^{2} and B=5x12B=\sqrt5 x^{-\frac{1}{2}}.
Use the rule amanam+na^{m} \cdot a^{n} \equiv a^{m+n} to help you simplify!
(edited 6 years ago)
I think the others are making (b) sound more complicated than it is. I'm assuming you won't have even covered logs yet given what the next question is.

Write sqrt(3) as 3^0.5, and write 81 as a power of 3. You can then use the rules of indices to simplify this so that you can directly compare it to the left hand side
(edited 6 years ago)
Reply 5
Original post by BobLoblawLawBlog
I think the others are making (b) sound more complicated than it is. I'm assuming you won't have even covered logs yet given what the next question is.

Write sqrt(3) as 3^0.5, and write 81 as a power of 3. You can then use the rules of indices to simplify this so that you can directly compare it to the left hand side


I agree, that would probably be the easiest way of solving it!
Reply 6
Original post by BobLoblawLawBlog
I think the others are making (b) sound more complicated than it is. I'm assuming you won't have even covered logs yet given what the next question is.

Write sqrt(3) as 3^0.5, and write 81 as a power of 3. You can then use the rules of indices to simplify this so that you can directly compare it to the left hand side


I ended up getting x = -6, is this correct?
Original post by iloading22
I ended up getting x = -6, is this correct?


Spoiler

Original post by iloading22
I ended up getting x = -6, is this correct?


Try substituting it back in and tell me if both sides then give the same answer.
Original post by iloading22
I ended up getting x = -6, is this correct?


Close but no.

I just realised as the user above said you may have not covered logs yet, you can put both sides to a base of 3, I'll show you what BobLoblawLawBlog means. Left hand side is already done.

So, 38131234\frac{\sqrt 3}{81} \Rightarrow \frac{3^{\frac{1}{2}}}{3^{4}}.I've just simplified the numerator and denominator to a base of 3.

Therefore 30.5x1=312343^{0.5x-1} = \frac{3^{\frac{1}{2}}}{3^{4}}. Simplify the right hand side then you can cancel the 3 out from both sides leaving you with a simple equation to solve for xx
(edited 6 years ago)
Reply 10
Original post by BobLoblawLawBlog
Try substituting it back in and tell me if both sides then give the same answer.


After rechecking it turns out I made a mistake and got x=-5 instead. Is this right?
Reply 11
Original post by ManLike007
Part b, multiply both sides by lnln or loglog. Use the law ln(a)bbln(a)ln(a)^{b} \equiv b \cdot ln(a) to help you simplify and rearrange.

Part a, if A=15x5A=\frac{1}{5}x^{5} then A2=(15x5)2A^{2} = (\frac{1}{5}x^{5})^{2}. Use the rule (ab)nanbn(ab)^{n} \equiv a^{n}b^{n}

Part b, (AB)2(AB)2(15x55x12)(AB)^{2} \equiv (A \cdot B)^{2} \Rightarrow (\frac{1}{5}x^{5} \cdot \sqrt5 x^{-\frac{1}{2}}). I've just substituted in A and B, that's all because you know that A=15x5A=\frac{1}{5}x^{5} and B=5x12B=\sqrt5 x^{-\frac{1}{2}}.
Use the rule amanam+na^{m} \cdot a^{n} \equiv a^{m+n} to help you simplify!


For part A i got : 1/25 x^4
For part B i got 1/5 x^3
Original post by iloading22
For part A i got : 1/25 x^4
For part B i got 1/5 x^3


That's correct and the first part is correct as well (x=5)(x=-5).

(I just noticed I wrote A=15x5A=\frac{1}{5}x^{5} when it's A=15x2A=\frac{1}{5}x^{2} from my message so I apologise)
(edited 6 years ago)
Original post by iloading22
After rechecking it turns out I made a mistake and got x=-5 instead. Is this right?


That sounds much better. Well done :smile:

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