Clement Cheng's answer to James's Malaysia question.
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Hope this helps.
I'm not sure if you wrote log2(2x-5) or log2(2x) - 5 though...
But I assume it's log2(2x-5)
I'm not sure if you wrote log2(2x-5) or log2(2x) - 5 though...
But I assume it's log2(2x-5)
Date Posted:
6 years ago
Thank you for solve this problem and explain it to me
U really help me a lot.i hope you can help me when I got any questions
No problem. Glad I was able to help.
I would like to ask something.why i need to change the base to base 2
This is to make the right hand side into a number, which is easier to work with.
You can also write log8(2) = log8(8^(1/3))
= (1/3)log8(8) = 1/3
= (1/3)log8(8) = 1/3
Am I need to follow the base of the log inside the bracket in the left side
What do you mean?
I mean do I need to follow the base 2 in the log 27(log base 2 ^2x-5)in the left side when I change the base in the right side
Can you show me that step more detailly
No, you don't have to follow. But for this question, it just happens that the base for the right hand side also changes to 2.
Oh
Clement Cheng I have another similar question.can I send it to you
Sure
Log16(log3^2x-1)=log4^2
The 2nd and 3rd log terms don't have bases?
Can I clarify if it's log(base 3)(2x - 1) or log(base3)(2x) -1 ?
Got.the second log base is 3 and third log is 4.sorry .I forget to write base there.
The first statement
Log(base 3)(2x-1)=log (base4)(2)
Ok thanks
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