SOLUTION: Hello, I'd like to know how to solve the following Logarithmic equation for x log<sub>125</sub>(1/5) = x I ended up doing this= 5·5^3x=5^1/

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Hello, I'd like to know how to solve the following Logarithmic equation for x log<sub>125</sub>(1/5) = x I ended up doing this= 5·5^3x=5^1/      Log On


   



Question 1090060: Hello, I'd like to know how to solve the following Logarithmic
equation for x
log125(1/5) = x


I ended up doing this= 5·5^3x=5^1/2
But it does not match with the solution I got on the calculator which is -6.
Thanks for your time.

Found 3 solutions by Edwin McCravy, Alan3354, MathTherapy:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
log%28125%2C%281%2F5%29%29%22%22=%22%22x

ANY logarithmic equation logB(A) = C
is equivalent to the exponential equation A = BC.

Therefore your logarithmic equation is equivalent to
the exponential equation 

1%2F5%22%22=%22%22125%5Ex

Next we write 1/5 as 5-1 and 125 as (53)

5%5E%28-1%29%22%22=%22%22%285%5E3%29%5Ex

We multiply the exponents on the right to remove the
parentheses:

5%5E%28-1%29%22%22=%22%225%5E%283x%29

Now Since the bases of the exponents on the laft and right are
equal, positive, and different from 1, the exponents must also
be equal, so we can drop the bases and set the exponents equal:

-1%22%22=%22%223x

Divide both sides by 3

-1%2F3%22%22=%22%22x

Edwin

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Makes no sense.
----------------
Logarithms are not solved, equations are solved.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Hello, I'd like to know how to solve the following Logarithm:
Log 125 (1:5)= X


I ended up doing this= 5·5^3x=5^1/2
But it does not match with the solution I got on the calculator which is -6.
Thanks for your time.
I don't know how you got what you got, but if it's matrix%281%2C3%2C+log+%28125%2C+%281%2F5%29%29%2C+%22=%22%2C+x%29, then: highlight_green%28matrix%281%2C3%2C+x%2C+%22=%22%2C+-+1%2F3%29%29