b. If 2° = 5= 10' prove that 1/a+1/b = 1/c
Answers
Answer:
Let,
{2}^{a} = {5}^{b} = {10}^{c} = k2
a
=5
b
=10
c
=k
Now,
The things we can get by using logarithm here is,
(a)
\begin{gathered}log_{2}(k) = a \\ \frac{ log(k) }{ log(2) } = a \\ \\ \frac{ log(2) }{ log(k) } = \frac{1}{a}\end{gathered}
log
2
(k)=a
log(2)
log(k)
=a
log(k)
log(2)
=
a
1
And,
(b)
\begin{gathered}log_{5}(k) = b \\ \frac{ log(k) }{ log(5) } = b \\ \frac{ log(5) }{ log(k) } = \frac{1}{b}\end{gathered}
log
5
(k)=b
log(5)
log(k)
=b
log(k)
log(5)
=
b
1
Similarly,
(c)
\begin{gathered}log_{10}(k) = c \\ \frac{ log(k) }{ log(10) } = c \\ \frac{ log(10) }{ log(k) } = \frac{1}{c}\end{gathered}
log
10
(k)=c
log(10)
log(k)
=c
log(k)
log(10)
=
c
1
Adding the first two equations gives,
\begin{gathered}\frac{ log(2) + log(5) }{ log(k) } \\\end{gathered}
log(k)
log(2)+log(5)
I'm gonna use an identity here where we have,
log(a) + log(b) = log(ab)log(a)+log(b)=log(ab)
Using this,
we get,
\begin{gathered}\frac{ log(10) }{ log(k) } \\\end{gathered}
log(k)
log(10)
Which is equal to (c).
Therefore,
\begin{gathered}\frac{1}{a} + \frac{1}{b} = \frac{1}{c} \\\end{gathered}
a
1
+
b
1
=
c
1
Cheers!