Question

show that:
log (base a)x/log (base ab)x=1+log (base a)b

Answer 1

did you try using change of base formula?

Answer 2

ya..i did but somehow its nt comng

Answer 3

\[\log_a(x)=\frac{\ln(x)}{\ln(a)}, \log_{ab}(x)=\frac{\ln(x)}{\ln(ab)}\]

Answer 4

\[\frac{\ln(x)}{\ln(a)} \div \frac{\ln(x)}{\ln(ab)}=\frac{\ln(x)}{\ln(a)} \cdot \frac{\ln(ab)}{\ln(x)}\]

Answer 5

now ask yourself does something cancel?

Answer 6

and then also can you rewrite ln(ab)

Answer 7

my subscription

Answer 8

are you still stuck?

Answer 9

yA PLEASE if posible mak it a litle clear...

Answer 10

\[\text{Let } k =log_{ab}(x)\]\[\implies ab^k = x\]\[\implies log_a\left((ab)^k\right) = log_a(x)\]\[\implies log_a(a^k) + log_a(b^k) = log_a(x)\]\[\implies k + k(log_a(b)) = log_a(x)\]\[\implies \frac{log_a(x)}{k} = 1 + log_a(b)\]
Substituting back in for k...
\[\implies \frac{log_a(x)}{log_{ab}(x)} = 1 + log_a(b)\]

Answer 11

Fun with logs!

Answer 12

Err that first line should be \[(ab)^k\]

Answer 13

Or the second.. whatever.

Answer 14

realllllllllyyyyyyy it helped me

Answer 15

thnk u

Answer 16

Just be sure you understand what was done at each step.

Answer 17

If not, please ask.

Answer 18

ya ...i got it..

Answer 19

\[\log_{a} x \log_{b} y \log_{c} z =\log_{b} x \log_{c} y \log_{a} z\]

Answer 20

prove

Answer 21

myininaya there is a factor by grouping problem waiting for you...