Question on exponentials and logarithms

Question

Given that $101^a=103^b=10403^c$, find an expression, in terms of $a$ and $b$, for $c$.

Answer 1

If $p^a=q^b=(pq)^c$

Let each of them $=k$

One of the values of $p$ is $k^{1/a}$

Similarly find $q,pq$ in terms of $k$

Now use $pq=p\cdot q$

$k^{1/c}=k^{1/a+1/b}$

In general $k\ne0,\pm1$

Can you take it from here?

Here $10403=101\cdot103 =pq$

Answer 2

Hint:

Given that $101^a=103^b=10403^c$, by taking the logarithm of each part in the equality we have:

$$\log(101^a)=\log(103^b)=\log(10403^c)$$

Now, recall how logs interact with exponents on their arguments

$\log(101^a)=a\cdot log(101)$, an exponent inside can be brought down as a coefficient outside.

Answer 3

Logging the equations gives us $\log(101^a)=\log(103^b)=\log(10403^c)$ which, using the logarithm properties, can be rewritten as $$a\log(101)=b\log(103)=c\log(10403).$$ Let us consider the equation $b\log(103)=c\log(10403)$. We can isolate for $c$ and write it as $$ c=\frac{b\log(103)}{\log(10403)}. $$ Note that since $a\log(101)=b\log(103)$ then $2b\log(103)=a\log(101)+b\log(103)$ meaning that $b\log(103)=\frac{1}{2}\left(a\log(101)+b\log(103)\right)$. With this fact, we can express $c$ in terms of $a$ and $b$ as $$ \begin{align} c&=\frac{1}{\log(10403)}\left(\frac{a\log(101)+b\log(103)}{2}\right)\\ &=\boxed{\frac{a\log(101)+b\log(103)}{2\log(10403)}.} \end{align} $$