Solve:
4^(x-1) = 7
2007-10-23 06:47:05 · 11 answers · asked by pearlnruby89 1 in Science & Mathematics ➔ Mathematics
log 4^(x - 1) = log 7
(x - 1) log 4 = log 7
x - 1 = log 7 / log 4
x = 1 + ( log 7 / log 4 )
Note
log can be to any base.
2007-10-23 19:54:25 · answer #1 · answered by Como 7 · 0⤊ 0⤋
Take the log of both sides; since the (x-1) is an exponent it becomes multiply:
(x-1) log 4 = log 7
Now using your calculator, divide log 7 by log 4, which would equal (x-1), then add 1
2007-10-23 13:51:42 · answer #2 · answered by hayharbr 7 · 0⤊ 0⤋
x=3
2007-10-23 13:49:11 · answer #3 · answered by Bing Bong Bao 3 · 0⤊ 2⤋
(x-1) * log 4 = log 7
x = log 7 / log 4 + 1
x= 2.4036
2007-10-23 13:51:27 · answer #4 · answered by tanzer360 5 · 0⤊ 0⤋
is that 4 raised to the (x-1)power or 4 multiplied by (x-1) If you add this information I will gladly answer the question
2007-10-23 13:52:21 · answer #5 · answered by ws 3 · 0⤊ 0⤋
x = 1 + log_base_4(7) = 1 + ln(7)/ln(4)
2007-10-23 13:50:39 · answer #6 · answered by Alina 2 · 0⤊ 0⤋
4^(x-1) = 7, so
ln 7 = (x-1) ln 4
x-1 = ln 7/ln 4 = 1.4037 (approx.)
x = 2.4037
2007-10-23 13:52:06 · answer #7 · answered by Mark H 3 · 0⤊ 0⤋
Kirk to Enterprise
2007-10-23 13:50:27 · answer #8 · answered by Anonymous · 0⤊ 1⤋
4^(x-1) = 7
LN4^(x-1) = (x-1)LN4 = LN7
x = LN7/LN4 +1 = 2.403677
2007-10-23 13:51:03 · answer #9 · answered by fcas80 7 · 0⤊ 0⤋
sori not so good in maths
2007-10-23 13:49:56 · answer #10 · answered by Perfectionist 6 · 0⤊ 1⤋