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C2 Logs question help

Hello

I'm stuck on this question, see attached. I have tried to do the working out and ended up with one correct answer (1) but there is another solution I missed (0). There must be an issue with my method - can anyone show me the correct way to approach this question and similar questions?

It's question 7e on the second picture
(edited 8 years ago)
image.jpg427.7KB
image.jpg511.4KB
Edit: didn't attach my working

image.jpg
Original post by h8skoooooool
Hello

I'm stuck on this question, see attached. I have tried to do the working out and ended up with one correct answer (1) but there is another solution I missed (0). There must be an issue with my method - can anyone show me the correct way to approach this question and similar questions?

It's question 7e on the second picture


It would help if you posted your method so we can see where you've gone wrong
Original post by h8skoooooool
Hello

I'm stuck on this question, see attached. I have tried to do the working out and ended up with one correct answer (1) but there is another solution I missed (0). There must be an issue with my method - can anyone show me the correct way to approach this question and similar questions?

It's question 7e on the second picture


easy you sub 5x 5^x as y instead
(edited 8 years ago)
Original post by samb1234
It would help if you posted your method so we can see where you've gone wrong


Sorry, it didn't attach - see above
Reply 5
Avatar for Zacken
Zacken
22
You're meant to notice that 52x=(5x)25^{2x} = (5^x)^2 so you have a quadratic in 5x5^x.

Make the substitution u=5xu = 5^x to get u2−6u−5=0u^2 - 6u - 5 = 0 then solve for u and back-substitute.
Original post by h8skoooooool
Edit: didn't attach my working

image.jpg


Use a substitution and see if that could help us
Original post by thefatone
easy you sub 5x 5^x as y instead


Oh so y^2 - 6y + 5 = 0 ?

I feel like an idiot now haha, I was going to do that but it didn't seem right
(edited 8 years ago)
Reply 8
Avatar for Zacken
Zacken
22
Original post by thefatone
easy you sub 5x 5^x as y instead
edit: 52x=5x×52\displaystyle 5^{2x} = 5^x \times 5^2


That's incorrect. 5x×52=5x+2≠52x5^x \times 5^2 = 5^{x+2} \neq 5^{2x}
Reply 9
Avatar for Zacken
Zacken
22
Original post by h8skoooooool
Oh so y^x - 6y + 5 = 0 ?

I feel like an idiot now haha, I was going to do that but it didn't seem right


Nah, y^2 not y^x. I assume it's just a typo though. :smile:
Original post by Zacken
That's incorrect. 5x×52=5x+2≠52x5^x \times 5^2 = 5^{x+2} \neq 5^{2x}


yes i realised xD i deleted now xD
Original post by Zacken
Nah, y^2 not y^x. I assume it's just a typo though. :smile:


Yeah sorry it was, thanks a lot!

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