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I HATE logarithms...

I am doing my P2 retake in Novemeber... after getting an unimpressive 69% in June last year... Once again I have stumbled upon my arch nemesis, logarithms! Those nonsensical and pointless fellows of the exponential function and demons catering to such a devils whim... Now I am more angry than ever with them... and am going to defeat them in battle.

PLEASE can someone show me the full method for how to answer these questions from Heinemann P2. Pg 113:

16. 3^(2x + 1) = 3^x + 24

18. 4^2x + 48 = 4^(x + 2)

21. Find the values of x for which
log3x - 2 logx3 = 1

22. Solve the equation:
log3(2 - 3x) = log9(6x^2 - 19x + 2)

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Reply 1
RobbieC
I am doing my P2 retake in Novemeber... after getting an unimpressive 69% in June last year... Once again I have stumbled upon my arch nemesis, logarithms! Those nonsensical and pointless fellows of the exponential function and demons catering to such a devils whim... Now I am more angry than ever with them... and am going to defeat them in battle.

PLEASE can someone show me the full method for how to answer these questions from Heinemann P2. Pg 113:

16. 3^(2x + 1) = 3^x + 24

18. 4^2x + 48 = 4^(x + 2)

21. Find the values of x for which
log3x - 2 logx3 = 1

22. Solve the equation:
log3(2 - 3x) = log9(6x^2 - 19x + 2)


16.
The method for these, generally, is to try and turn them into quadratics:

Let y = 3^x

3^(2x + 1) = 3^x + 24

3.3^2x = 3^x + 24

3y^2 = y + 24

3y^2 - y - 24 = 0

(3y + 8)(y - 3 ) = 0

y = -8/3 OR 3

Since y = 3^x; taking logs:

ln(y) = xln(3)

x = [ln(y)]/[ln(3)]

Since we can only take logs of positive numbers, we can ignore y = -8/3 and just concentrate on y = 3.

ln(y) = ln(3)

Hence, x = 1.

The second one is very similar.

Ben
Reply 2
Thanks, I didnt know how to handle the 2x + 1 in the brackets...

Could anyone do the last two I mentioned... Im finding it a task!!! Book once again has no relevant examples... Good ol' rubbish Heinemann.
Reply 3
RobbieC
21. Find the values of x for which log3x - 2 logx3 = 1


for question 21 does x=9?
Reply 4
Find the values of x for which
log3x - 2 logx3 = 1

log(3) x - 2 log(x) 3 = 1
log(3) x - 2[log(3) 3]/[log(3) x] = 1
log(3) x - 2/[log(3) x] = 1
(log(3) x)^2 - log(3) x - 2 = 0
(log(3) x - 2)(log(3) x + 1) = 0

log(3) x = 2
3^2 = x
x = 9

log(3) x = -1
3^-1 = x
x = 1/3

you need to know the laws very well so you can always form a quadratic when you need to. substituting a = log(3) x will help make it clearer.

I used the law:

log(a) x = [log(b) a]/[log(b) x]

to convert the base from x to 3.
Reply 5
meepmeep
For 22, note that 3^2=9, so log9(6x^2 - 19x + 2)=log3(12x^2-38x+4), so we get:

log3(2 - 3x) =log3(12x^2-38x+4). So 2-3x=12x^2-38x+4 which is solvable more easily.

?

im confused?

using change of base formula you dont get that.

log9(6x^2 - 19x + 2)
=[log3(6x^2 - 19x + 2)] / [log39]
=[log3(6x^2 - 19x + 2)] / 2
=[log3(6x^2 - 19x + 2)^0.5]

?
Reply 6
we did 22 in class and I found it easier to square the base and the (2-3x) from the left hand side and simply get rid of the logs and solve for x like a normal quadratic
Reply 7
mik1a
...
I used the law:

log(a) x = [log(b) a]/[log(b) x]

to convert the base from x to 3.

Actually, thar should be the other way around.#

log(a) x = [log(b) x]/[log(b) a]

check it out with,

log(10) 5 = ln(5)/ln(10)
Reply 8
kikzen
?

im confused?

using change of base formula you dont get that.

log9(6x^2 - 19x + 2)
=[log3(6x^2 - 19x + 2)] / [log39]
=[log3(6x^2 - 19x + 2)] / 2
=[log3(6x^2 - 19x + 2)^0.5]

?



Oops. What on earth am I playing at? You're completely right. Just ignore what I wrote. Had a brainstorm...
Reply 9
Im so stupid... I cant do 23, though I did the others with your guys' help... If someone could do this one, id be most grateful... Last piece of the logarithmic puzzle:
2^2x+1 = 3(2^x) - 1
Reply 10
RobbieC
Im so stupid... I cant do 23, though I did the others with your guys' help... If someone could do this one, id be most grateful... Last piece of the logarithmic puzzle:
2^2x+1 = 3(2^x) - 1


Is that:
2^(2x+1) = 3(2^x) - 1
?
Reply 11
RobbieC
Im so stupid... I cant do 23, though I did the others with your guys' help... If someone could do this one, id be most grateful... Last piece of the logarithmic puzzle:
2^2x+1 = 3(2^x) - 1

Let y = 2^x, form quadratic, solve for y and, finally, solve for x.

I get x = 0 and x = -1 as solutions.

Ben
Reply 12
DAMN! I am an idiot. Sorry about that... I just removed the +1 bit from the first exponential and got a y^2 term with the wrong coefficient... then tried to factorise using the formula and it went downhill.. Thanks for the help.
Reply 13
RobbieC
DAMN! I am an idiot. Sorry about that... I just removed the +1 bit from the first exponential and got a y^2 term with the wrong coefficient... then tried to factorise using the formula and it went downhill.. Thanks for the help.


You solved it with minimal help. Seems you aren't an idiot to me.
Reply 14
I would rep you ben, but ive run out for today... PM me, and ill be sure to rep you tomorrow.

-Rob
Reply 15
Gaz031
You solved it with minimal help. Seems you aren't an idiot to me.
Nice of you to say, but being self-critical helps me to improve. It's a little mad, but I kinda feel I have to do it... especially when living around people who won't criticise me...

What's weird is that I don't like criticism from others, unless I ask for it... Yes... I am deranged.
Reply 16
Don't wrorry, P2 June 2004 was a beast
Reply 17
hihihihi
Don't wrorry, P2 June 2004 was a beast

really? :tongue: :biggrin:
Reply 18
mockel
really? :tongue: :biggrin:

heh it was probably a breeze for you geniuses

I remember it started off with a simplification question which was so easy you could do it in one line as everything cancelled. Then it had the some of the hardest questions for each section.

edexcel maths btw.
Reply 19
hihihihi
heh it was probably a breeze for you geniuses

I remember it started off with a simplification question which was so easy you could do it in one line as everything cancelled. Then it had the some of the hardest questions for each section.

edexcel maths btw.

yeah, i kinda remember sitting that paper