Operation of transistor in Opamp log amplifier

[brainbaby]

In the figure it could also be possible to connect the transistor Q1 as a diode by shorting its collector with its base ..but instead it is done by keeping both base and collector at ground...
The text have to say that if it would be connected the other way(i.e by shorting collector and base )then the base current would have caused an error ( Base current error ) .
Which type of error is the author talking about..?
Why and how does the error manifest itself...?

Please help!

 

[Svein]

The circuit around the first OpAmp uses the exponential relationship between V_be and I_c in the feedback to create an output V_o = -C⋅log(V_in) - V_beo. The circuit around the second OpAmp removes V_beo and amplifies the log output.

If you look at Q1, you see that I_b goes from ground (constant potential) into the OpAmp output (very low output resistance) and therefore is not part of the collector current. If collector and base were shorted, the base current and the collector current would be added.