loga(7/8)
(a) Rewrite the logarithm as a ratio of common (log_10) logarithms

Question

anonymous · Answer

loga(b/c)=loga(b)-loga(c)=log10(b)/log10(a)-log10(c)/log10(a).
here b=7 and c=8.

anonymous · Answer

i got caught up wit hthe a...

anonymous · Answer

brackett, is tive me -0.057

anonymous · Answer

give *

anonymous · Answer

i got the result right?

anonymous · Answer

yeah...for calculatin the answer we need the value for a.

anonymous · Answer

the final result is?

anonymous · Answer

log10(7)/log10(a) -log10(8)/log10(a)

anonymous · Answer

$$log_a(\frac{7}{8})=\frac{log(\frac{7}{8})}{log(a)}$$

anonymous · Answer

yes worldboy

anonymous · Answer

loga( 7/8) = loga(7) – loga(8)
                  =log10(7)/log10(a) -log10(8)/log10(a)
in agreement with brackett.
since we don't know anything about log 10(a) neither part can be simplified further.

anonymous · Answer

my apologies for a typing error earlier.  I put loga(16x) when it should have been loga(7/8)