VCE Methods Question Thread!

[Bluegirl]

Hey bluegirl,

try and think about what equals, if you broke it up. It corresponds with one of the log laws.

Got it, thanks! Seemed so obvious once I looked at it like that haha.

Express a in terms of b if :
Log 10(a)= 2-b
No idea where to start.

How do I change 2 into a log?

[Nato]

Got it, thanks! Seemed so obvious once I looked at it like that haha.

Express a in terms of b if :
Log 10(a)= 2-b
No idea where to start.

How do I change 2 into a log?

change 2 into a log? i'm not exactly sure what you mean by that.
so with we want to express in terms of . well our log laws come into play again.
hint, it's this one: then

[Bluegirl]

change 2 into a log? i'm not exactly sure what you mean by that.
so with we want to express in terms of . well our log laws come into play again.
hint, it's this one: then

I don't know either.

So, log 10^2-b= a
I don't know

[noah the lettuce]

here are a few suggestions
for 1. start by letting then solve for a. After you receive the value(s) for , make it equal to , to solve for

for 2. try and use the factor theorem, so subbing in values that make equal to . This will help you in finding a factor. Which you can then use long division to find other factors, and hence the solution(s)

for 3. rearrange the equation to look like: . Now you can square both sides of the equation to 'eliminate' the square root sign. After a bit of expanding and rearranging you will be left with a 'normal' quadratic, which you can then solve for

for 4. kinda similar to 3. you could start off by squaring both sides:
so on the right hand side the square root will be eliminated, and on the left hand side, i would advise to expand it (using difference of perfect squares). this expansion will resulting in a square root term, but after a bit of rearranging and squaring both sides, you can find the value(s) for x.

Let me know, if that didn't make sense haha cheers!

We'll give you hints rather than doing all the working for you.
1. Multiply both sides by to get rid of the you have in one of the denominators.
2. Group the terms and , take a factor out of the first which should leave you with both groupings that have another factor in common, then factorise this factor out to give you something that you can work with via the null factor law.
3. You need to first get rid of the square root, so to do this we normally square both sides, but as we have , if you square the left side you'll still get a square root when you expand it. So we first need to get the square root on it's own by moving the across, then square both sides and expand out to give you a quadratic. ALso take note of any domain restrictions that arise from having the square root in there (remember anything 'underneath' the square root has to be equal to or greater than zero, which sometimes restricts your solutions).
4. You'll need to do the rearranging and squaring both sides twice for this one, as the first time you'll still end up with a square root left over. Again be careful with domain restrictions from the square root.

Spoiler

4.

Check domain restrictions,

Which means that is not a valid solution.

EDIT: Beaten.

Wow, thanks a million times, they definably helped me work them out, I was stuck on all of those and had tried to work them out before but kept getting wrong answers so cheers!! Also had no idea about domain restrictions on the square roots   

[soNasty]

How would I simplify into ?

This takes too long to write on an iPad....

[Nato]

I don't know either.

So, log 10^2-b= a
I don't know

you're close, bluegirl. You just didn't need to include the 'log'. following the log law, the answer would be

[Nato]

How would I simplify into ?

This takes too long to write on an iPad....

what you can do is multiply the top and bottom of by the conjugate of the denominator (which is ). You are essentially, multiply the fraction by one.This will aid in getting rid of any square roots signs in the denominator.

[b^3]

How would I simplify into ?

This takes too long to write on an iPad....

You want to rationalise the denominator by multiplying by the conjugate over the conjugate of the denominator. That is if we have then we would multiply by , effectively you flip the sign. As we are multiplying by something over itself, we are effectively multiplying by 1.

Spoiler

EDIT: Beaten.

[Bluegirl]

you're close, bluegirl. You just didn't need to include the 'log'. following the log law, the answer would be

That's not the answer though. I got the rest with a friends help. Thankyou

[Nato]

That's not the answer though. I got the rest with a friends help. Thankyou

out of curiosity, what was the answer?

[soNasty]

Thank you so much!

[Bluegirl]

out of curiosity, what was the answer?

a= 100 x 0.1^b

[alchemy]

a= 100 x 0.1^b

If anyone is else wondering, this is how you get the answer:
log 10(a)= 2-b
a=10^(2-b)
a=(10^-b)*(10^2)
a=100*(1/10)^b
a=100*0.1^b

Nato was close, but he wasn't *exactly* expressing in terms of .

[Blondie21]

Shouldn't the answer be ?

If anyone is else wondering, this is how you get the answer:
log 10(a)= 2-b
a=10^(2-b)
a=(10^-b)*(10^2)
a=100*(1/10)^b
a=100*0.1^b

Nato was close, but he wasn't *exactly* expressing in terms of .

Could the answer have also been ??
Does it equal the same thing or am I just doing illegal math

[soNasty]

how would i show that cos(100) is equivalent to 2cos^2(50)-1?
i know this question is using that double angle formula thingy..

is cos(100) [degrees] in the form of cos(2theta) already? meaning that theta is equal to 50? so just subbing it in to the appropriate formula will give u the abovementioned answer? or am i on the wrong track.. haha