SOLUTION: Log base 5 (r+2) + log base 5 (r-2) = 1
log base 3 (a-3) = 1 + log base 3 (a+1)
log exponent 2x = (log x exponent 2
Thanks
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Question 294482: Log base 5 (r+2) + log base 5 (r-2) = 1
log base 3 (a-3) = 1 + log base 3 (a+1)
log exponent 2x = (log x exponent 2
Thanks
Answer by unlockmath(1688) (Show Source): You can put this solution on YOUR website!
Hello,
I'll help you out with the first one (Log base 5 (r+2) + log base 5 (r-2) = 1)
This can be rewritten as:
Log base 5 (r+2)(r-2) = 1
Now let's exponentiate both sides by using base 5 to get:
5^log5(r+2)(r-2)=5^1 This can then be written as:
(r+2)(r-2)=5 Expand this out to get:
r^2-4=5 Add 4 to both sides to get:
r^2=9 Square root both sides will result in:
r=+-3
Make sense?
RJ
www.math-unlock.com
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