a level logs help

hollyeb3

log11(2x-1)=1-log11(x+4)

I have no idea where to start for this. any help would be appreciated

Reply 1

Ryanzmw

Original post by hollyeb3

log11(2x-1)=1-log11(x+4)

I have no idea where to start for this. any help would be appreciated

To get you started:

\log_{11}(2x-1) + \log_{11}(x+4) = \log_{11}((2x-1)(x+4)) = 1

and remember that:

\log_{a}(y) = 1 \iff y = a

Reply 2

brainmaster

Original post by hollyeb3

log11(2x-1)=1-log11(x+4)

I have no idea where to start for this. any help would be appreciated

bring logs on one side and simplify;
log 11 (2x-1) + log 11 (x+4) = 1
log 11 (2x-1)(x+4) = 1
11^1 = (2x-1)(x+4).

hope u can do it now :smile:

Reply 3

EnterUsername.

just a q, how you you change to base of any log to 10 or something else?

Reply 4

Prasiortle

Original post by EnterUsername.

just a q, how you you change to base of any log to 10 or something else?

The change-of-base formula gives

\log_{a}{b} = \frac{\log_{c}{b}}{\log_{c}{a}}

- if you haven't seen this before, try to prove it for yourself! It will help to rewrite using exponentials rather than logarithms.

Reply 5

HashMash

Original post by brainmaster

bring logs on one side and simplify;
log 11 (2x-1) + log 11 (x+4) = 1
log 11 (2x-1)(x+4) = 1
11^1 = (2x-1)(x+4).

hope u can do it now :smile:

What would the answer be though?
I got 11=2x^2+7x-4
So then I did a quadratic and got 2 anwsers obviously. Now im confused what to do. Can someone help

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