Question

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

Answer 1

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

Answer 2

i got caught up wit hthe a...

Answer 3

brackett, is tive me -0.057

Answer 4

give *

Answer 5

i got the result right?

Answer 6

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

Answer 7

the final result is?

Answer 8

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

Answer 9

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

Answer 10

yes worldboy

Answer 11

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.

Answer 12

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