Asked by confused--two questions please help
solve the following logarithmic equation.
1. 2log(lower2) (x-2) + log(lower2)^2=3
I do not know how to make the 2 by the g on my computer.
2. log(lower729^x) + log(lower27^x) + log(lower3^x) =9
again I do not know how to make the numbers lower case
please show work, and if you can tell me how to make the lower case numbers on here.
1. 2log(lower2) (x-2) + log(lower2)^2=3
I do not know how to make the 2 by the g on my computer.
2. log(lower729^x) + log(lower27^x) + log(lower3^x) =9
again I do not know how to make the numbers lower case
please show work, and if you can tell me how to make the lower case numbers on here.
Answers
Answered by
Steve
lots of folks just write log_2(x-2)
the numbers are subscripts, not lower-case.
1. since we're dealing with logs base 2, just say log:
2log(x-2) + log2 = 3
log(x-2)^2 + log2 = 3
log(2(x-2)^2) = 3
2(x-2)^2 = 8
(x-2)^2 = 4
x = 0 or 4
I assumed that log(lower2)^2 meant log(lower2)2
2.
since 729 = 3^5, log base 729 = 1/5 log base 3
since 27 = 3^3, log base 27 = 1/3 log base 3
so, using all logs base 3,
1/6 logx + 1/3 logx + logx = 9
3/2 logx = 9
logx = 6
x = 3^6 = 729
the numbers are subscripts, not lower-case.
1. since we're dealing with logs base 2, just say log:
2log(x-2) + log2 = 3
log(x-2)^2 + log2 = 3
log(2(x-2)^2) = 3
2(x-2)^2 = 8
(x-2)^2 = 4
x = 0 or 4
I assumed that log(lower2)^2 meant log(lower2)2
2.
since 729 = 3^5, log base 729 = 1/5 log base 3
since 27 = 3^3, log base 27 = 1/3 log base 3
so, using all logs base 3,
1/6 logx + 1/3 logx + logx = 9
3/2 logx = 9
logx = 6
x = 3^6 = 729
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