SOLUTION: Can you help me with some questions to help me understand how to do it?
log5 7x=2
log3(n-3)=4
logx(49)=2
log(1000)=x
log2(2^5^X^-1)=9
log(4x-1)=2
Question 1031659: Can you help me with some questions to help me understand how to do it?
log5 7x=2
log3(n-3)=4
logx(49)=2
log(1000)=x
log2(2^5^X^-1)=9
log(4x-1)=2 Found 2 solutions by ikleyn, josmiceli:Answer by ikleyn(52792) (Show Source): You can put this solution on YOUR website! .
Can you help me with some questions to help me understand how to do it?
log5 7x=2 7x =
log3(n-3)=4 n-3 =
logx(49)=2 x = 7
log(1000)=x = 3 ---> x = 3.
log2(2^5^X^-1)=9 ambiguous since you don't use parentheses properly
log(4x-1)=2 = 2 ---> 4x-1 = 100.
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website! Use inverse functions
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( note that base 10 is assumed when
the base is not specified )
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( hope I copied this right )
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You may have trouble "reading" the
log functions
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In general, if you have:
log ( to some base ) ( that gives me this result ) = the log, which is an exponent
or, with symbols:
The left side is TELLING you that the right side is a log ( exponent )
You know the base is , so the equation has to look like:
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You can go in either direction, too
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Then you have to apply the laws of logs and
the inverse of the log -the exponent function
Hope this helps