Help with logarithm problem?

2 ⤊

Help with logarithm problem?

Simply put:

Find x in:
10^x+3 = 6^2x

2007-12-08 18:00:51 · 2 answers · asked by N.M.M.P.P.I. 2 in Science & Mathematics ➔ Mathematics

2 answers

10^(x+3) = 6^2x
(x+3) log 10 = 2x log6
x log 10 + 3 log 10 = 2x log 6
x log 10 - 2x log 6 = -3 log 10
x (log 10 - 2 log 6) = -3 log 10
x = -3 log 10 / (log 10 - 2 log 6)
x = -3 / (1 - 2 log 6)
x = 5.39275

Best of luck! :)

2007-12-08 18:12:35 · answer #1 · answered by disposable_hero_too 6 · 1⤊ 0⤋

Assume question is meant to read as:-
10^(x + 3) = 6^(2x)
Take log base ten:-
(x + 3) = log 6^(2x)
x + 3 = 2x log 6
x + 3 = 2x (0.778)
x + 3 = 1.556 x
3 = 0.556 x
x = 5.4 (to 1 decimal place)

2007-12-09 07:47:22 · answer #2 · answered by Como 7 · 1⤊ 0⤋