Question

Hello! I need to find\[\log _{5}6\] when\[\lg2=a\] and \[\lg3=b\]

Answer 1

Assuming "lg" is also used for log, and its base is also 5, you can use the rule:
\[\log_{b}p+\log_{b}q=\log_{b}pq\]By setting a=5, p=2 and q=3.

Answer 2

well, I have always thought that

\[lga=\log _{10}a\]
But thanks, anyway!

Answer 3

You could be right.
In that case, \(log_{5}6=\dfrac{lg6}{lg5}=\dfrac{lg2+lg3}{lg5}=\dfrac{a+b}{lg5}\).
I used the rule \(log_{b}a=\dfrac{log_{c}a}{log_{c}b}\), to change the base from 5 to 10.

Answer 4

oh,thank you a lot!

Answer 5

the answer is

\[\frac{ a+b }{ 1-a }\] maybe do you know how they get a-1?

Answer 6

oh, *1-a, my bad!

Answer 7

@mathslover any ideas?:/

Answer 8

wait!

Answer 9

btw lg means "log _ 2 "

Answer 10

I searched net about it

Answer 11

oh sorry \(lg = \log_{10} \)

Answer 12

well, I don know what lg means, we learnt about it at school. But i dont know how to get 1-a :o

Answer 13

I do know*

Answer 14

wait I am getting it

Answer 15

okayy!

Answer 16

what is log 10 (base 5) ?

Answer 17

10^x=5?

Answer 18

see what I did :
\[\large{\log_{10} 2 + \log_{10} 3 = \log_{10}(6)}\]
\[\large{a + b = \log _{10}(6)}\]
Changing base :
\[\large{\log_{10}(6) = \frac{\log_5(6)}{\log_5{10}}}\]
\[\large{(a+b)(\log_5(10))= \log_5(6)}\]

Answer 19

gtg now, sorry I will discuss it later

Answer 20

\[ \log_{5}(6)= \log_{5}(2 \cdot 3)=\log_{5}(2)+\log_{5}(3)\]
now use the rule
\[ \log_{a}(x)= \log_{a}(b)\log_{b}(x) \]
with a=5, b=10 and x=3 we get
\[ \log_{5}(3)= \log_{5}(10)\log_{10}(3) \]
write 10 as 2*10 so we get
\[ \log_{5}(3)= (\log_{5}(2)+ \log_{5}(5))\log_{10}(3) \]
use the given
\[ \log_{10}(3)=b \]
and
\[ \log_{5}(5)=1 \]
to get
\[ \log_{5}(3)= b\log_{5}(2)+ b \]

now let's look at the other term
\[ \log_{5}(2) = \log_{5}(10)\log_{10}(2) \]
\[ \log_{5}(2) = (\log_{5}(2)+ \log_{5}(5))\log_{10}(2) \]
use the given
\[ \log_{10}(2)=a \]
\[ \log_{5}(2) = a\log_{5}(2)+ a\]
solve for \( \log_{5}(2) \)
\[ \log_{5}(2)= \frac{a}{1-a} \]

we now know enough to get the final answer

Answer 21

Picking it up again from where I left:

\(\log_{5}6=\dfrac{a+b}{\lg 5}\) (still assuming lg is log base 10).
\(\log_{5}6=\dfrac{a+b}{\lg 5+\lg2-\lg2}\)
\(\log_{5}6=\dfrac{a+b}{\lg(5 \cdot2)-\lg2}\)
\(\log_{5}6=\dfrac{a+b}{\lg 10-\lg2}\)
So finally we get:
\(\log_{5}6=\dfrac{a+b}{1-a}\)

Answer 22

\[ \log_{5}(2)+\log_{5}(3)= \frac{a}{1-a} +\frac{ab}{1-a}+ b \]
which simplifies to your answer

Answer 23

thank you all!!