SOLUTION: How do you solve this problem. And please explain because I am having trouble.
log x = 2 log a 3 + log a 5
To do this I just first switched it so that log x = 2 would be at
Algebra ->
Exponential-and-logarithmic-functions
-> SOLUTION: How do you solve this problem. And please explain because I am having trouble.
log x = 2 log a 3 + log a 5
To do this I just first switched it so that log x = 2 would be at
Log On
Question 416151: How do you solve this problem. And please explain because I am having trouble.
log x = 2 log a 3 + log a 5
To do this I just first switched it so that log x = 2 would be at the end instead of the beginning.
And then I used the first law of logarithm and got this:
2loga^15=logaX
Now what? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! log x = 2 log a 3 + log a 5
:
Assuming the problem is:
log(x) = 2log(a^3) + log(a^5)
which is
log(x) = 3*2log(a) + 5log(a)
log(x) = 6log(a) + 5log(a)
log(x) = 11*log(a)
which is
log(x) = log(a^11)
therefore
x = a^11
:
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