Can Complex Logarithms Challenge Advanced Math Students?

[jspen30]

Hello

I am currently studying Math B in high school and am having an extremely hard time with my current assessment. If anyone could please help me with the equations listed below it would be very much appreciated as my current Math teacher is on long service leave, and our substitute is practically teacher herself at the same time as teaching our class.Equation 1:

Log 50 - 2 Log 5 - Log 2Equation 2:

Log (x-3) = 1 + Log 4 - Log xEquation 3:

2^-4a + 2 x 2^-2a - 8 = 0Equation 4:

Without using calculator, find the value of A that makes x = e^12 / 1-2e^12 a solution of the equation:

ln x - ln (ax+1) = 12
I have attempted the first few on my own, how ever have not found the correct answer when checked against that of my calculator. They are listed below

Equation 1:

y=Log 50 - 2 Log 5 - Log 2
= 2 Log (50/5/2)
= 2 Log 5

Answer should be in Index form however I cannot make this into index formI'm sorry for such a long post, and don't expect to be given the questions directly, however any help would be appreciated, even if it just pointing me in the right direction or explaining how to complete them. I'm very grateful for any help you may provide

Thankyou

Jake

 

[Dick]

Now what rules did you use to change Log 50 - 2 Log 5 - Log 2 into 2 Log 5? Can you spell it out?

 

[jspen30]

i believe i used the law

Log_a xy = log_a x + log_ay

However I am not one hundred percent sure as my math teacher is currently away on stress leave (I doubt it was my class), and therefore I may be missing some formulae or equations to compete these questions

Thanks again for your quick reply

Jake

 

[Dick]

i believe i used the law

Log_a xy = log_a x + log_ay

However I am not one hundred percent sure as my math teacher is currently away on stress leave (I doubt it was my class), and therefore I may be missing some formulae or equations to compete these questions

Thanks again for your quick reply

Jake

What did you do with the '2' in front of '2 Log 5'? Use a*log(b)=log(b^a).

 

[jspen30]

it was placed out the front eg

y=Log 50 - 2 Log 5 - Log 2
= 2Log (50/5/2)
= 2 Log 5

however this is not the final answer, as that is what I am trying to find out, the final answer should be in index form

im trying to work out how i would do this equation to end up with an answer in index form, or if there is someway of going from the answer

2 Log 5 --- to index form

Thanks

 

[Dick]

it was placed out the front eg

y=Log 50 - 2 Log 5 - Log 2
= 2Log (50/5/2)
= 2 Log 5

however this is not the final answer, as that is what I am trying to find out, the final answer should be in index form

Here's an example. 1+2*3+4 is not equal to 2*(1+3+4). You can't pull the 2 on the second term out in front of the expression. That's an algebra mistake.

 

[jspen30]

ok thankyou for telling me this mistake, How would I be able to find the following, showing full working and without using a calculator?

y = Log 50 - 2 Log 5 - Log 2
= 2 Log (50/5/2)
= 2 Log 5

Thanks again

Jake

 

[Dick]

ok thankyou for telling me this mistake, How would I be able to find the following, showing full working and without using a calculator?

y = Log 50 - 2 Log 5 - Log 2
= 2 Log (50/5/2)
= 2 Log 5

Thanks again

Jake

Use a*log(b)=log(b^a). So 2*log(5)=log(5^2), right? Now apply your other law.

 

[jspen30]

ok thankyou

my current working out is:

Log 50 - Log 5^2 - Log 2
Log (50 / 5^2 / 2)
=1

Is this correct?

Jake

 

[Dick]

ok thankyou

my current working out is:

Log 50 - Log 5^2 - Log 2
Log (50 / 5^2 / 2)
=1

Is this correct?

Jake

You mean Log(1), right? That's not 1.

 

[jspen30]

yes I am very sorry

Log 50 - Log 5^2 - Log 2
Log (50 / 5^2 / 2)
= Log 1
= 0

is that correct working?

 

[Dick]

yes I am very sorry

Log 50 - Log 5^2 - Log 2
Log (50 / 5^2 / 2)
= Log 1
= 0

is that correct working?

Yes, it is. You don't usually want to write something like 50/25/2, though I know what you mean. Best to use some parentheses. (50/25)/2=1, 50/(25/2)=4. Not the same. The first one is the correct way to write it. Or 50/(25*2).

 

[jspen30]

ok ill re-adjust my notation, thanks again for all the help, you have just taught me in about 5mins what my current substitute couldn't teach my class in 5 lessons. Do you have the knowledge to help me with any of the other equations listed above. If you don't have a lot of spare time or have more important things to do I understand.

Thanks

Jake

 

[Dick]

ok ill re-adjust my notation, thanks again for all the help, you have just taught me in about 5mins what my current substitute couldn't teach my class in 5 lessons. Do you have the knowledge to help me with any of the other equations listed above. If you don't have a lot of spare time or have more important things to do I understand.

Thanks

Jake

The most important thing I have to do right now is go to sleep. But I'll get you started. 'Log' with any indicated base often means log to the base e. If that's the case then take e to the power of both sides. Use e^(Log(a))=a.

 

[jspen30]

ok thanks for all your help, and sorry for keeping you up (I live in Australia, and its only 2pm)

Cheers

Jake

 

[Dick]

You aren't keeping me up. I'm keeping myself up. I'm sure somebody else can give you some help if I'm not around. Keep working on them and show your attempts.

 

[Redbelly98]

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