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#1 Re: Help Me ! » Simplify the following: » 2014-12-03 02:38:52

I have made the amendment- thank you

#3 Re: Help Me ! » Simplify the following: » 2014-12-01 21:08:01

Hi;

I am given it a try concerning the conversion;

Log x = 1/2log x

so, it will be
log (x) = 1/2log (x)

Is that correct?

#4 Re: Help Me ! » Simplify the following: » 2014-12-01 06:38:30

Please, help.

I don't understand the question.

#5 Re: Help Me ! » Simplify the following: » 2014-12-01 06:24:24

Show that
Log x = 1/2logx and hence solve log x + 1/2 = logx


In both cases the LHS is in base 4 and the RHS is in base 2


Thanks

#6 Re: Help Me ! » Simplify the following: » 2014-11-24 21:37:32

This is it;

log(x^2 + 1) - 2logx = 1

#7 Re: Help Me ! » Simplify the following: » 2014-11-24 21:22:41

bobbym wrote:

Ask the tutor to plug and check his answer.


Yes - have seen the tutor - the tutor gave me a calculator which I plugged 1/3 into the equation and I had 1(one),  so it seems his answer is correct. What do you say Bobbym?

#8 Re: Help Me ! » Simplify the following: » 2014-10-25 21:27:08

I have plug it in and the calculator gave me "math error". But I want to add that the base is five [5] for all.

#9 Re: Help Me ! » Simplify the following: » 2014-10-25 05:46:49

Ok. Let me put this

Log(x-3) - logx = 2.
Is negative one (-1) the correct answer?

#10 Re: Help Me ! » Simplify the following: » 2014-10-25 05:04:27

Good!
Thanks bobym - I will post problems.

Big thanks!

#14 Re: Help Me ! » Simplify the following: » 2014-10-25 04:26:23

xlog16 = log10
xlog16 = 1
x = 1/log16
Correct?

#15 Re: Help Me ! » Simplify the following: » 2014-10-25 04:17:20

Thanks! Bobbym

I am now getting logarithm better.
See;
solve for x in,  log16^x.

log16^x = 10
xlog16 = log10
x = log10/log16
correct?

#16 Re: Help Me ! » Simplify the following: » 2014-10-25 04:03:01

I think your methods  are good concerning the simplification of like terms in logs. Thanks for answering.

Look it if it is good
log16^-x = log2 + log7
-xlog16 = log14
  x= -log14/log16
Is it good?

#17 Re: Help Me ! » Simplify the following: » 2014-10-25 03:46:52

Still I haven't come across the tutor - but I will definitely meet him.

My one question is - when do one take antilogs of both sides of an equation?

#18 Re: Help Me ! » Simplify the following: » 2014-10-21 04:54:01

I am away from the tutor now - but I shall tell him later

#19 Re: Help Me ! » Simplify the following: » 2014-10-21 04:27:00

The book says - if the base is not written then it is base 10. And with this problem the base is not written - so it used base 10

#20 Re: Help Me ! » Simplify the following: » 2014-10-21 04:14:47

Wow!
I confronted a math tutor - but he claimed the answer 1/3 or -1/3 is correct - now I Could realise that your method is good. Please I want the tutor to see for himself your steps because he was insisting that the books' answer is correct

Thanks bobbym

#21 Re: Help Me ! » Simplify the following: » 2014-10-21 03:58:49

Yes that's the problem - I copied it correctly.

When you plug in your answers does it give 1?

#23 Re: Help Me ! » Simplify the following: » 2014-10-20 04:55:26

Hi, Bobbym, I was called away, I am sorry - but I don't have my calculator with me now.
It is negative or positive 1/3 in the book as an answer. When I get to  the house I will punch them in the calculator and see if it really gives negative or positive 1/3.

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