Question

Does anyone know how to solve Base 10 logarithm problems?

Answer 1

what problem r u facing in solving it?

Answer 2

yeah

Answer 3

18^(x-2) = 13^(-2x)
Taking logarithm to the base 10,
(x-2) log 18 = -2x log 13
or x log 18 + 2x log 13 = 2 log 18
or (log 18 + 2 log 13) x = 2 log 18
or x = (2 log 18)/(log 18 + 2 log 13)

Answer 4

Probably

Answer 5

\[~~~~~~~~~~~~~18^{(x-2)} = 13^{(-2x)}\]\[~~~~~log_{10}18^{(x-2)} = log_{10}13^{(-2x)}\]\[(x-2)log_{10}18 = (-2x)log_{10}13\]\[(x-2)log_{10}18 + (2x)log_{10}13=0\]\[xlog_{10}18 + 2xlog_{10}13=2log_{10}18 \]\[x(log_{10}18+2log_{10}13)=2log_{10}18 \]\[x={2log_{10}18 \over log_{10}18+2log_{10}13}\]

all are correct