Sum of exponents equivalent to card(composites less than or equal to x)? Consider a cousin of the Chebyshev function:  
$$t(x) = \sum \alpha_i $$ such that  $$ p_i^{\alpha_i }= x, \ p_i \leq x$$
I speculated that  $ t(x) \sim C(x)$, the composites $\leq(x)$
If $t(x) = \sum \alpha_i = \sum \log_{p_i}x  = \sum \frac{\ln x}{\ln p_i } = \ln x \sum \frac{1}{\ln p_i}\sim \ln x $ Li(x) $  \sim \ln x \frac{x}{ \ln  x} \sim x \sim C(x) $. 
After some thought the above seems okay, and also for 
$$r(x) = \sum \beta_j $$ such that  $$c_j^{\beta_j }= x, \ c_j \leq x$$ in which $c_j$ is the jth composite (excluding 1). 
Because s = $\sum\frac{1}{\ln c_j} $ is nearly $\sum \frac{1}{\ln x}$ we would expect that $\ln x\sum \frac{1}{\ln c_j}\sim x$ also.
My lingering question is this:  why would 
$\sum \beta_j - x \sim \pi(x)$, or equivalently, $\ln x\sum \frac{1}{\ln c_j}- x \sim \pi(x)$?
A typical calculation: x= 900,000, r(x) = 971141, r(x) - x = 71141 and $$\pi(x) = 71274$$ 
 A: First, I wanted to point out that most of what you wrote is not correct.  $t(x)\sim\frac{x}{\log x}$, not $\sim x$.  This is because if $\sum_{p\leq x}\sim \text{li(x)}$, then we expect that by dividing by $\log p$, I will get $\text{li}(x)/\log(x)$, and then $t(x)\sim \text{li}(x)$.  Also, the computations are way off.  Checking on Matlab, I have that:  $$r(900000)=9000090.94\dots$$ and hence $$r(900000)-900000=90.94\dots$$   You did not remove the primes correctly.  Now lets prove all of this in full.
Some Rigorous Proofs:  Lets go over everything in detail.
The two series you are considering are
$$t(x)=\log x\sum_{p\leq x}\frac{1}{\log p}$$ and $$r(x)=\log x\sum_{1<c\leq x}\frac{1}{\log c},\ \text{where }c>1\text{ is composite.}$$ Notice that $r(x)+t(x)=\sum_{2\leq n\leq x}\frac{1}{\log n}$ so we need only evaluate $\sum_{2\leq n\leq x}\frac{1}{\log n}$  and $\sum_{p\leq x}\frac{1}{\log p}$. First
Lemma 1: $$\sum_{2\leq n\leq x}\frac{1}{\log n}=\text{li}(x)-C+O\left(\frac{1}{\log x}\right)$$ where the constant $C$ is given by $$C=\int_{2}^{\infty}\frac{\left\{ t\right\} }{t\left(\log t\right)^{2}}dt.$$ 
Proof: We may write this as a Riemann-Stieltjes integral: $$\sum_{2\leq n\leq x}\frac{1}{\log n}=\int_{2}^{x}\frac{1}{\log t}d\left[t\right]=\int_{2}^{x}\frac{1}{\log t}dt-\int_{2}^{x}\frac{1}{\log t}d\left\{ t\right\}$$  where $\left[t\right]$  and $\left\{ t\right\}$   are the floor and fractional parts of $t$, respectively. Using integration by parts and the definition of $\text{li}(x)$  this is $$\text{li}(x)-\frac{\left\{ x\right\} }{\log x}-\int_{2}^{x}\frac{\left\{ t\right\} }{\left(\log t\right)^{2}}dt.$$ 
Remark: Although I have not done the computation myself, I believe the constant $C$ can be cleaned up in terms of other known constants.
Lemma 2: We have that $$\sum_{p\leq x}\frac{1}{\log p}=\text{li}\left(x\right)-\frac{x}{\log x}+O\left(xe^{-c\sqrt{\log x}}\right).$$ 
Proof: If $\theta(x)=\sum_{p\leq x}\log p$, we may write  $$\sum_{p\leq x}\frac{1}{\log p}=\int_{2}^{x}\frac{1}{\left(\log t\right)^{2}}d\theta\left(t\right)=\int_{2}^{x}\frac{1}{\left(\log t\right)^{2}}dt+\int_{2}^{x}\frac{1}{\left(\log t\right)^{2}}d\left(\theta(t)-t\right).$$ For the first term, notice that by integration by parts  $$\int_{2}^{x}\frac{1}{\log t}dt=\frac{x}{\log x}+\int_{2}^{x}\frac{1}{\left(\log t\right)^{2}}dt.$$ For the second term we can use integration by parts along with the prime number theorem which states that $\theta(t)-t=O\left(xe^{-c\sqrt{\log x}}\right).$ Then $$\int_{2}^{x}\frac{1}{\left(\log t\right)^{2}}d\left(\theta(t)-t\right)=\frac{\left(\theta(x)-x\right)}{\left(\log x\right)^{2}}+2\int_{2}^{x}\frac{\theta(t)-t}{t\left(\log t\right)^{3}}dt+O(1)$$  $$=O\left(xe^{-c\sqrt{\log x}}\right),$$ and the lemma follows.
Consequences:  Putting these two lemmas together, we find that: $$t(x)=\text{li}(x)\log x-x+O\left(xe^{-c\sqrt{\log x}}\right),$$
$$r(x)+t(x)=\text{li}(x)\log x -C\log x+O\left(xe^{-c\sqrt{\log x}}\right).$$  We have to put in the extra error term since we are subtracting $r(x)$, and this will consume the $C\log x$ term since it is larger.  Hence $$r(x)=x+O\left(xe^{-c\sqrt{\log x}}\right)$$ and 
$$r(x)-x=O\left(xe^{-c\sqrt{\log x}}\right).$$
I hope that helps,
