Q:

$\log(n+1) = 1 + 4/3\log(c(n+1))$ solve for c

$$\log(n+1) = 1 + 4/3\log(c(n+1))$$
how do I solve for c so that this equation holds? 
$$\log(n+1) = 1 + \log(c(n+1)) = 1 + 4/3\log(c) + 4/3\log(n+1)$$
now what?

A:

$$\log(n+1)= 1 + \frac{4}{3}\log c + \frac{4}{3} \log(n+1)$$
$$\Leftrightarrow$$
$$1-\frac{1}{3}\log(n+1) = -\frac{4}{3}\log c $$
$$\Leftrightarrow$$
$$3 - \log(n+1) = -4\log c$$
$$\Leftrightarrow$$
$$4\log(n+1) -12 = \log c$$
$$c = e^{4\log(n+1)-12}$$
The final expression can be further simplified.

