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[gnome]

Would someone please show me why

{ a^{log_cb} = b^{log_ca}

for all a, b and c.

[deltabourne]

I don't know if this is really a proof.. but just take the log of both sides:
{ a^{log_cb} = b^{log_ca}
log_c { a^{log_cb} = log_c b^{log_ca}
({log_cb})({log_ca}) = ({log_ca})({log_cb})
using the property that:
{log_ca^r} = r{log_ca}

[matt grime]

it is a proof. perhaps to make it appear more rigorous, you could write a^{log_cb}=c^{log_c(a^{log_cb})} and similarly for the rhs, and say that the only way for c^xto be equal to c^y is if x=y.

[gnome]

Got it!

Thanks, folks.