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

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> 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      Log On


   

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(52790) About Me  (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 = 5%5E2

log3(n-3)=4          n-3 = 3%5E4

logx(49)=2           x = 7

log(1000)=x          log%2810%2C%281000%29%29 = 3  --->  x = 3.

log2(2^5^X^-1)=9     ambiguous since you don't use parentheses properly

log(4x-1)=2          log%2810%2C%284x-1%29%29 = 2  --->  4x-1 = 100.


Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Use inverse functions
+log%28+5%2C+7x+%29+=+2+
+5%5E2+=+7x+
+7x+=+25+
+x+=+25%2F7+
--------------------
+log%28+3%2C+n-3+%29+=+4+
+3%5E4+=+n-3+
+n-3+=+81+
+n+=+84+
---------------------
+log%28+x%2C49+%29+=+2+
+x%5E2+=+49+
+x%5E2+=+7%5E2+
+x+=+sqrt%28+7%5E2+%29+
+x+=+7+
---------------------
+log%28+1000+%29+=+x+
( note that base 10 is assumed when
the base is not specified )
+10%5Ex+=+1000+
+10%5Ex+=+10%5E3+
+x+=+3+
---------------------
+log%28+2%2C+%28%282%5E5%29%5Ex%29%5E%28-1%29+%29+=+9+
( hope I copied this right )
+log%28+2%2C+2%5E%28-5x%29+%29=+9+
+2%5E9+=+2%5E%28+-5x+%29+
+9+=+-5x+
+x+=+-9%2F5+
-------------------------
+log%28+4x+-+1+%29+=+2+
+10%5E2+=+4x+-+1+
+4x+=+10%5E2+%2B+1+
+4x+=+101+
+x+=+25.25+
-------------------------
You may have trouble "reading" the
log functions
--------------------------
In general, if you have:
log ( to some base ) ( that gives me this result ) = the log, which is an exponent
or, with symbols:
+log%28+a%2C+b+%29+=+c+
The left side is TELLING you that the right side is a log ( exponent )
You know the base is +a+, so the equation has to look like:
+a%5Ec+=+b+
----------------
You can go in either direction, too
+a%5Ec+=+b+
+log%28+a%2Cb+%29+=+c+
------------------
Then you have to apply the laws of logs and
the inverse of the log -the exponent function
Hope this helps