If log base 2 (3) + log base 3 (4) + ...... log base x (x+1) = 7, what is x?

We have the sum of logarithms with different bases:
\log_{2}{3} + \log_{3}{4} + ...... + \log_{x}{(x+1)} = 7

We can use the change of base formula to write each logarithm in base 10:
\log_{2}{3} + \log_{3}{4} + ...... + \log_{x}{(x+1)} = \frac{\log{(3)}}{\log{(2)}} + \frac{\log{(4)}}{\log{(3)}} + ...... + \frac{\log{(x+1)}}{\log{(x)}} = 7

Since the logarithms are equal to 7, we have:
\frac{\log{3}}{\log{2}} + \frac{\log{4}}{\log{3}} + ...... + \frac{\log{(x+1)}}{\log{x}} = 7

Since \log{a} - \log{b} = \log{\left(\frac{a}{b}\right)} , we can simplify the expression:
\log_{2}{3} + \log_{3}{4} + ...... + \log_{x}{(x+1)} = \log_{2}{\frac{3}{2}} + \log_{3}{\frac{4}{3}} + ...... + \log_{x}{\frac{(x+1)}{x}} = 7

By the properties of logarithms, we have:
\log_2{\frac{3}{2} \cdot \frac{4}{3} \cdot ...... \cdot \frac{(x+1)}{x}} = 7

Simplifying the expression inside the logarithm:
\log_2{\frac{(x+1)}{2}} = 7

To solve for x:
\frac{(x+1)}{2} = 2^7
(x+1) = 2^8
x = 2^8 - 1 = 255

Therefore, the value of x is \boxed{255} .