C2 Logarithms

I was revising Logs and I thought a good way would be to make my own questions, it makes the method easier to recall for me. But I made the following equation:

2^(4x) = 19 + 3^(x)

And it left me stumped! I was wondering if anyone who fancied showing off their log skills or just wants more revision for tommorow could show me how? Thanks :smile:

Reply 1

ak9779

You won't be asked that in a C2 exam. I've got C4 coming up and have no idea if that's even possible... My best guess would be rewriting 19+3^x as one exponential term otherwise you're going nowhere...

(edited 12 years ago)

Reply 2

Tom722

Hahaha ok, I'm actually retaking C2 but the knowledge from C3 really helps the understanding! But I did check it on Wolfram|Alpha and it's roughly x = 1.12 so it must be possible. Anyway thanks for pointing that out, and good luck for C4!

what does the ^ mean

would be much, much easier if instead of 3^x you'd have 2^x
try that one instead :smile:

Reply 5

Sashari

Original post by an obese donut

what does the ^ mean

to the power of

Reply 6

an obese donut

i got x=0.29, im probably wrong but i took logs of both sides, brought the powers down in front of the log, rearranged to get x on one side, then factorise x out and rearrange again and work out

Reply 7

nuodai

Original post by an obese donut

i got x=0.29, im probably wrong but i took logs of both sides, brought the powers down in front of the log, rearranged to get x on one side, then factorise x out and rearrange again and work out

I don't know what you did but it definitely wasn't right (just try subbing it in).

The only reason why the standard tricks aren't working here is because of the pesky 19. You can't say that

\log (A+B) = \log A + \log B

, for example, so it's impossible to split it up into something you can take logs of easily. I think there's a typo in the question -- it can't be solved algebraically, as far as I can see.

Reply 8

jwza

I took logs of everything and brought the powers down to get
4xlog2=log19+xlog3
then got the x's on one side and took x out to get
x(4log2-log3)=log19
then divided by the bracket to find x which i got to be 1.76

tbh i didnt think that was wrong but as everyones saying it cant be done it must be

Reply 9

an obese donut

Original post by jwza

thats what i did but just put it into the calc wrong

Reply 10

an obese donut

but its still not right

Reply 11

jwza

Original post by an obese donut

but its still not right

How? Its bugging me that i cant find whats up with it :tongue:

Reply 12

Moonlight19

Lolol, I got x = 26.13 (to 2 d.p.) :P
Definitely wrong :P

Reply 13

Moonlight19

Original post by jwza

19 is not log 19
19 = 19log(10) (or so i thought?)

Reply 14

j.alexanderh

Original post by jwza

In bold is the mistake. log(19+3^x) =/= log19 + xlog3.

log(A+B) =/= logA + logB

Reply 15

chou246

Original post by jwza

The underlined bit's wrong -

+19 in logs would be + 19log10

Wolfram alpha seems to be able to do it, so it can be done.

Reply 16

an obese donut

this is certainly one hard log

Reply 17

jwza

Original post by chou246

...

Original post by j.alexanderh

...

Original post by Moonlight19

...

Ah yeah i forgot that. thanks

Reply 18

ak9779

Someone said that the calculator can do this one - I expect they use some form of iterative method rather than an algebraic one and so it remains impossible without a machine...

Reply 19

Tom722

Oh my I feel a bit guilty causing such a fuss! And even worse because the log questions were nothing in comparision to this! Haha, again thank you all for the trouble and good luck in the exams :smile: