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AS Logarithms problem - C2

Logarithms problem. Edexcel examboard, C2.

Find to three significant figures, the value of x for which:

3^{2x+1} = 105

I solved it in this way.

3^{2x+1} = 105[br][br]log(3^{2x+1}) = log(105)[br][br]2x+1 log(3) = log(105)[br][br]2x+1 = \dfrac{log(105)}{log(3)}[br][br]x = (\dfrac{log(105)}{log(3)} \div 2) - 1[br][br]x = 1.118...

However, when I put my solved value of x back into the equation

3^{2x+1} = 105

I get the answer 35, not 105?

Please could someone tell me where I am going wrong?

Scroll to see replies

Reply 1

Ben121

You made an error on your fifth line. Think about what happens when you divide through by 2 first.

Reply 2

Wilko94

Original post by Ben121

You made an error on your fifth line. Think about what happens when you divide through by 2 first.

I still don't know what I've done wrong, sorry.

Reply 3

Nomolos

( \frac{log(105)}{log(3)} \div 2 ) - 1

should be

( \frac{log(105)}{log(3)} -1 ) \div 2

Reply 4

jameswhughes

Subtract 1 then divide by 2, not the other way round.

Reply 5

Wilko94

Original post by Wilko94

I still don't know what I've done wrong, sorry.

Ok - I've realised I need to divide by 2 after I minus the 1.

Please could you explain to me why?

Reply 6

Wilko94

Original post by jameswhughes

Subtract 1 then divide by 2, not the other way round.

I did just recognise that before you posted, but thank you anyway. Please could you explain to me why it works that way?

Reply 7

jameswhughes

Original post by Wilko94

Ok - I've realised I need to divide by 2 after I minus the 1.

Please could you explain to me why?

You are trying to find x from 2x+1, so if you divide by 2 first you end up with x+0.5, not x or x+1

Reply 8

Nomolos

It's simple order of method,
for example..
if you had

2x+1 = 5

you would subtract 1 from both sides before dividing by two, to give you

x = 2

. If you were to divide by two first, you would have an incorrect answer of

x = 1.5

. If you insist on dividing first, you must divide the left hand side completely by 2, leaving

x + \frac{1}{2} = 2.5

followed by the subtraction giving the correct answer of

x = 2

Reply 9

Wilko94

Original post by jameswhughes

You are trying to find x from 2x+1, so if you divide by 2 first you end up with x+0.5, not x or x+1

Thank you very much.

Reply 10

Scottishsiv

I don't understand why you have to go the long way, you could just easily do

log3(105) = x (use calculator to find value, log is to the base 3) you should have a scientific calculator make the most of it :biggrin:

Then (x-1) divide by 2.

Wouldn't that be much easier? Thats how I did C2 Logs.

(edited 13 years ago)

Reply 11

Wilko94

Original post by Scottishsiv

Then (x-1) divide by 2.

Wouldn't that be much easier? Thats how I did C2 Logs.

I have a scientific calculator - how do I change my calculator to base 3 instead of base 10?

Reply 12

jameswhughes

Original post by Scottishsiv

Then (x-1) divide by 2.

Wouldn't that be much easier? Thats how I did C2 Logs.

My calculator is quite old, it can only do base 10 or natural logs.

Reply 13

Kareir

Hmm, well, you could use

Log3(105) = 2x + 1

And Log3(105) = Log10(105)/log10(3)?

Edit: Log conversion rule

When you get the answer, check this.

Spoiler

_Kar.

(edited 13 years ago)

Reply 14

Wilko94

I see, but that makes it just as long-winded as the other way IMO.

I have another problem I'm stuck on. :frown:

[br]2^{2x-1} - 2^x - 21 = 0[br]

My teacher told me to solve it by letting

2^x = y

, which gives me

2y^2 - y - 21

which I then factorised to

(2y-7)(y+3)

.

What do I do now? :confused:

Reply 15

jameswhughes

Original post by Wilko94

I see, but that makes it just as long-winded as the other way IMO.

I have another problem I'm stuck on. :frown:

[br]2^{2x-1} - 2^x - 21 = 0[br]

My teacher told me to solve it by letting

2^x = y

, which gives me

2y^2 - y - 21

which I then factorised to

(2y-7)(y+3)

.

What do I do now? :confused:

Solve the quadratic, so you get y=3.5 and y=3, then re-write y as 2^x, then take logs, so x=(log3.5)/(log2). The negative root is invalid as logs can only be taken of positive numbers.

Reply 16

Wilko94

Original post by jameswhughes

Solve the quadratic, so you get y=3.5 and y=3, then re-write y as 2^x, then take logs, so x=(log3.5)/(log2). The negative root is invalid as logs can only be taken of positive numbers.

Thanks - I didn't know about the negative thing. I would give you another +rep if I could.

Reply 17

jameswhughes

Oops, made a mistake in my previous comment. It should say y=-3. Thanks for the rep though :biggrin:

Reply 18

Scottishsiv

Original post by Wilko94

I have a scientific calculator - how do I change my calculator to base 3 instead of base 10?

if you have the casio calculator fx-82 or above (if you don't you really should buy one, they are really good, you can practically do anything mathematical on them and they are allowed in the exam) it is the button underneath the on button and input the numbers and you are all set