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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)

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)

Scroll to see replies

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)

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

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.

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.

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.

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]

?

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

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

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.

Gaz031

You solved it with minimal help. Seems you aren't an idiot to me.

What's weird is that I don't like criticism from others, unless I ask for it... Yes... I am deranged.

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.

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

- a level biology maths question
- Math Help
- Arrhenius equation help
- What A-level maths skills are required in A-level chemistry and biology?
- What youtubers or websites are great to start learning logoithmic functions? Maths
- Do you use logarithms, matrices and exponentials in algorithms?
- a level maths study materials
- Chemistry
- My grow your grades journal 🎀 ~ as a year 12
- A level physics or maths for medicine?
- goals for the week! a-levels edition 🤍
- logbase e or logbase 10?
- A Level Maths Differentiation Question
- Physics A-Level
- Math help - Logarthmic functions
- Econ help - Finding the market Equilibrium
- a level options
- -1^x = b is this solveable using natural logarithms?
- Can I do A level Physics ?
- Mechanical Engineering University of Bolton

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