2007-04-23 20:42:05 · 4 answers · asked by lykastar 3 in Science & Mathematics ➔ Mathematics
10^(2 - 2x) = 2*10^(-x)
Using base 10 logarithms,
(2 - 2x)log10 = log2 - xlog10
2 - 2x = log2 - x
2 - log2 = x
2007-04-23 20:50:10 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋
RHS, 10^ (2-2x) = 100 / [(10)^x] ^2
LHS, 2 * 10^(-x) = 2 / (10)^x
as we have (10)^x on both sides, we can just simplify the equations as follows:-
100 / (10)^x = 2
100 = 2 * (10)^x
applying 'log' on both sides,
log100 = log 2 + x log10
as we know that log100=2 and log10 = 1, hence,
2 = log 2 + x
x = 2 - log2
2007-04-23 20:53:41 · answer #2 · answered by eve88 2 · 0⤊ 0⤋
Logs are to base ten in this case:-
10^(2 - x) = 2
(2 - x).log10 = log 2
2.log 10 - log 2 = x.log 10
2 - log 2 = x
2007-04-23 21:44:13 · answer #3 · answered by Como 7 · 0⤊ 0⤋
go to bed
2007-04-23 20:44:53 · answer #4 · answered by USMarine Dad 3 · 0⤊ 0⤋