It says solve to correct 2 decimal places. I keep getting it wrong.
If it's not 1.27 or rounded 1.28 or .78 then what is it ?
log equations: 7(5)^x=94?
2016-06-11 3:01 pm
回答 (3)
2016-06-11 3:04 pm
5^x = 94/7
x log5 = log94 - log7
x = (log94 - log7) / log5, my calculator gives
x = 1.6138458111
x log5 = log94 - log7
x = (log94 - log7) / log5, my calculator gives
x = 1.6138458111
2016-06-14 10:46 pm
7(5)^x=94
5^x = 94/7
x = log5 94/7
x = (log94/7)/(log5)
The answer is... x = (log94/7)/(log5) or 1.61.
5^x = 94/7
x = log5 94/7
x = (log94/7)/(log5)
The answer is... x = (log94/7)/(log5) or 1.61.
2016-06-11 6:55 pm
5^x = 94/7
5^x = 13.428571428571428571428571428571
Take logs to base '10'
log(10)5^x = log(10)13.42857...
x log(10)5 = log(10)13.42857...
x = log(10)13.42857... / log(10)5
x = 1.1280298... / 0.69897...
x = 1.6138458
x = 1.61 ( 2 d.p.)
5^x = 13.428571428571428571428571428571
Take logs to base '10'
log(10)5^x = log(10)13.42857...
x log(10)5 = log(10)13.42857...
x = log(10)13.42857... / log(10)5
x = 1.1280298... / 0.69897...
x = 1.6138458
x = 1.61 ( 2 d.p.)
收錄日期: 2021-04-21 18:58:31
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