https://postimg.cc/7JVx6q24

not really sure on how to get the correct answer to this which is D.

at first I tried solving n^10 = n + 1 to get n = 1 but that was wrong.

I Dont understand the working for the given markscheme either. completely confused, any ideas?

not really sure on how to get the correct answer to this which is D.

at first I tried solving n^10 = n + 1 to get n = 1 but that was wrong.

I Dont understand the working for the given markscheme either. completely confused, any ideas?

Original post by MonoAno555

https://postimg.cc/7JVx6q24

not really sure on how to get the correct answer to this which is D.

at first I tried solving n^10 = n + 1 to get n = 1 but that was wrong.

I Dont understand the working for the given markscheme either. completely confused, any ideas?

not really sure on how to get the correct answer to this which is D.

at first I tried solving n^10 = n + 1 to get n = 1 but that was wrong.

I Dont understand the working for the given markscheme either. completely confused, any ideas?

See https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/

The change base formula

Would it help?

Original post by MonoAno555

https://postimg.cc/7JVx6q24

not really sure on how to get the correct answer to this which is D.

at first I tried solving n^10 = n + 1 to get n = 1 but that was wrong.

I Dont understand the working for the given markscheme either. completely confused, any ideas?

not really sure on how to get the correct answer to this which is D.

at first I tried solving n^10 = n + 1 to get n = 1 but that was wrong.

I Dont understand the working for the given markscheme either. completely confused, any ideas?

Hi, I have the answer if you want. Are you still having problems?

Original post by Hehgxhrhfh

Hi, I have the answer if you want. Are you still having problems?

Please be aware of the forum rules which state that you should not provide solutions - only hints!

Original post by davros

Please be aware of the forum rules which state that you should not provide solutions - only hints!

Original post by MonoAno555

not really sure on how to get the correct answer to this which is D.

at first I tried solving n^10 = n + 1 to get n = 1 but that was wrong.

I Dont understand the working for the given markscheme either. completely confused, any ideas?

could you also tell me where you got the problem from?

Original post by Hehgxhrhfh

could you also tell me where you got the problem from?

Ultimate ENGAA Collection (with over 400 questions and solutions) book pdf from 2018

Original post by MonoAno555

Ultimate ENGAA Collection (with over 400 questions and solutions) book pdf from 2018

Dont know how the model solution did it, but I presume it was using the change of base formula and cancelling/telescoping mentioned in #2? If so, a related way (if you forget the change of base formula) is to note its a product of logs so use the power rule

k*log(x) = log(x^k)

so for the first two you have

log_2(3)*log_3(4) = log_2(3^log_3(4)) = log_2(4)

with an obvious generalisation to the product of n-1 logs. As the power rule is used to derive the change of base formula, its a similar method.

(edited 10 months ago)

Original post by BankaiGintoki

See https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/

The change base formula

Would it help?

The change base formula

Would it help?

Ty for this, I forgot this method

Original post by mqb2766

Dont know how the model solution did it, but I presume it was using the change of base formula and cancelling/telescoping mentioned in #2? If so, a related way (if you forget the change of base formula) is to note its a product of logs so use the power rule

k*log(x) = log(x^k)

so for the first two you have

log_2(3)*log_3(4) = log_2(3^log_3(4)) = log_2(4)

with an obvious generalisation to the product of n-1 logs. As the power rule is used to derive the change of base formula, its a similar method.

k*log(x) = log(x^k)

so for the first two you have

log_2(3)*log_3(4) = log_2(3^log_3(4)) = log_2(4)

with an obvious generalisation to the product of n-1 logs. As the power rule is used to derive the change of base formula, its a similar method.

I got the answer actually with the change in base method. I was also not aware you could simply logs this way. Thanks

(edited 10 months ago)

Original post by MonoAno555

I got the answer actually with the change in base method. I was also not aware you could simply logs this way. Thanks

Its one way to prove the change of base formula as

log_a(b) = log_a(c^log_c(b)) = log_c(b) log_a(c)

so

log_c(b) = log_a(b) / log_a(c)

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