Prove that if $2^{k_1}l_1=2^{k_2}l_2$, then $k_1=k_2$ and $l_1=l_2$. Suppose $k_1,k_2,l_1,l_2$ are natural numbers such that $l_1$ and $l_1$ are odd. Prove that if $2^{k_1}l_1=2^{k_2}l_2$, then $k_1=k_2$ and $l_1=l_2$.




First, let $l_1 = 2m_1 +1$ and $l_2=2m_2+1$ where $m_1,m_2\in\mathbb{N}$.

$$\begin{align*}
2^{k_1}l_1&amp;=2^{k_2}l_2\\
\frac{l_2}{l_1}&amp;=2^{k_1-k_2}\\
\ln\left(\frac{l_2}{l_1}\right)&amp;=({k_1-k_2})\ln(2)\\
\ln(l_2)-\ln(l_1)&amp;=({k_1-k_2})\ln(2)\\
\ln(2m_2+1)-\ln(2m_1 +1)&amp;=({k_1-k_2})\ln(2)\\
\end{align*}$$



I am stuck at this step because I can't see how I can get RHS or LHS to $0$.

Can anyone give me a hit or suggestion to keep going or start to rewrite a new proof?

Thanks.
Suppose $k_1\neq k_2$. If $k_2 &gt; k_1$, then dividing by $2^{k_1}$ gives
$$ l_1 = 2^{k_2 - k_1} l_2. $$
As $k_2-k_1 &gt; 0$, we deduce that $l_1$ is even. Contradiction.
Similarly, if $k_2 &lt; k_1$, we obtain a contradiction.
So we conclude that $k_1 = k_2$  which clearly implies $l_1 = l_2$.
We note that an odd number divided by an odd number gives an odd number. We also note that $2^n$ for $n \not=0$ is always an even number. So, for $\frac{l_2}{l_1}=2^{k_1-k_2}$ to be true, both sides must be equal to 1, which is equivalent to $l_1=l_2$ and $k_1=k_2$.