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Help with questions C1 Surds

Can someone show some step by step solutions to these two questions

Thank You in advance.

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Full solutions are not to be shared on this forum.

I will get you started on Q7.

3^(2x+1) * 9^x = 27

3^(2x+1) * (3^2)^x = 27

3^(2x+1) * (3^2x)= 27

Can you use the law a^b * a^c = a^ (b+c) now?

Can you therefore solve for x subsequently using simple logarithms?

Peace.
You cant use logs its C1
Original post by WhiteGroupMaths
Full solutions are not to be shared on this forum.

I will get you started on Q7.

3^(2x+1) * 9^x = 27

3^(2x+1) * (3^2)^x = 27

3^(2x+1) * (3^2x)= 27

Can you use the law a^b * a^c = a^ (b+c) now?

Can you therefore solve for x subsequently using simple logarithms?

Peace.


You cant use logs its C1
Reply 4
Original post by bobjon22444
You cant use logs its C1


Yes you can.

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Maybe it depends on the exam board. But you don't learn logs until c2.
Original post by Kholmes1
Maybe it depends on the exam board. But you don't learn logs until c2.


The question is from an edexcel paper. So indices laws should be used not logs. Because for logs a calculator is required
Lol I just want to solutions. The answer to both is x=1/2 but I want to know how it got there, stupid books dont give solutions.
Original post by bobjon22444
The question is from an edexcel paper. So indices laws should be used not logs. Because for logs a calculator is required


Yeah thats what I thought as well. You can't use it in c1. Do you know what paper it is? Exam Solutions is good.
Reply 9
Original post by Kholmes1
Maybe it depends on the exam board. But you don't learn logs until c2.


Original post by bobjon22444
The question is from an edexcel paper. So indices laws should be used not logs. Because for logs a calculator is required


Oh yeah. I forgot they called them indices laws in C1 but it is the same principle

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Original post by WhiteGroupMaths
Full solutions are not to be shared on this forum.

I will get you started on Q7.

3^(2x+1) * 9^x = 27

3^(2x+1) * (3^2)^x = 27

3^(2x+1) * (3^2x)= 27

Can you use the law a^b * a^c = a^ (b+c) now?

Can you therefore solve for x subsequently using simple logarithms?

Peace.


Is the answer for part a, x= 1/2?
Its from a pearson book called revisons work book. I cant seem to find the solutions either.
These aren't surds...
Original post by zigocarn
Is the answer for part a, x= 1/2?


Yes but how did you get it. It must be without logs.
Original post by Hydeman
These aren't surds...


LOL no its index laws.
Original post by bobjon22444
LOL no its index laws.


Yes, which is why I'm wondering why the thread advertises it as a surd problem...
Original post by bobjon22444
Yes but how did you get it. It must be without logs.


It's simple.
What you want to do first is make the bases the same for all three values.
Once the bases are the same, you are able to add the powers.
Remember that the power for both sides are equal to each other and this is why you can solve for x.
Reply 17
Original post by bobjon22444
Yes but how did you get it. It must be without logs.


Make everything be in the form of 3 to the power something. For example 27 would be 333^3.

Then use the rule azay=az+ya^z*a^y=a^{z+y} to solve for x

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Original post by Andy98
Make everything be in the form of 3 to the power something. For example 27 would be 333^3.

Then use the rule azay=az+ya^z*a^y=a^{z+y} to solve for x

Posted from TSR Mobile


so 3^2x+1*3^2=3^3 then add powers ?
Reply 19
Original post by bobjon22444
so 3^2x+1*3^2=3^3 then add powers ?


:yep:

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