SOLUTION: This is my problem in logarithm log x + log x^2 + log x^3 + log x^4 = 1 + log 0.2 + log 0.03^2 + log 0.004^3 thank's

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: This is my problem in logarithm log x + log x^2 + log x^3 + log x^4 = 1 + log 0.2 + log 0.03^2 + log 0.004^3 thank's      Log On


   



Question 813675: This is my problem in logarithm
log x + log x^2 + log x^3 + log x^4 = 1 + log 0.2 + log 0.03^2 + log 0.004^3
thank's

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!

A general procedure for solving these kinds of equations:
  1. Use algebra and/or properties of logarithms to transform the equation into one of the following forms:
    • log(expression) = other-expression
    • log(expression) = log(other-expression) (Note: The bases of the two logs must match.)
  2. Eliminate the logarithms:
    • If the equation is in the first form, "log(expression) = other-expression", rewrite the equation in exponential form.
    • If the equation is in the second form, "log(expression) = log(other-expression)", set the arguments equal.
  3. Now that the logs are gone, solve the equation (using techniques which are appropriate for the type of equation it is).
  4. Check your solution. This is not optional! A check must be made to see if the bases and arguments of all logs are valid. Any "solution" which make any base or an argument invalid must be rejected! (Note: Valid bases are positive but not 1 and valid arguments are positive.)
Let's try this on your equation. First we decide which form we think will be easiest to achieve. With the "non-log" term of 1 (on the right side), it would seem that the second, "all-log" form will be harder to reach. So we will aim for the first form.

Stage 1: Transform
To reach this form, all we need to do is find a way to combine all the logs into a single logarithm. We will start by combining all the logs on each side into single logarithms. For this we will use the log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Aq%29%29 property:
log%28%28x%2Ax%5E2%2Ax%5E3%2Ax%5E4%29%29+=+1+%2B+log%28%280.2%2A0.03%5E2+%2A+0.004%5E3%29%29
Simplifying...
log%28%28x%5E10%29%29+=+1+%2B+log%28%280.2%2A0.0009+%2A+0.000000064%29%29
log%28%28x%5E10%29%29+=+1+%2B+log%28%280.00000001152%29%29
Now we will get the logs on the same side. Subtracting the log on the right we get:
log%28%28x%5E10%29%29+-+log%28%280.00000001152%29%29+=+1+
And now we can use another property of logarithms, log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29, to combine the remaining logs:
log%28%28x%5E10%2F0.00000001152%29%29+=+1+
We have now reached the first form.

Stage 2: Eliminate the logs.
With the first form we just rewrite the equation in exponential form. In general log%28a%2C+%28p%29%29+=+n is equivalent to p+=+a%5En. Using this pattern, and the fact that the base of "log" is 10, we get:
x%5E10%2F0.00000001152+=+10%5E1
which simplifies to:
x%5E10%2F0.00000001152+=+10

Srage 3: Solve
Our equation is now an exponential equation. But with only one x term it will not be hard to solve. Multiplying each side by 0.00000001152 we get:
x%5E10+=+0.0000001152
Now we find the 10th root of each side (remembering both the positive and negative roots):
x+=+root%2810%2C+0.0000001152%29 or x+=+-root%2810%2C+0.0000001152%29

Stage 4: Check
Use the original equation to check:

Checking x = x+=+root%2810%2C+0.0000001152%29:
This can be done by inspection. This value of x is positive. So all the powers of x on the left side will be positive if x is positive. So all the bases arguments are valid. So this value checks out!
Checking x+=+-root%2810%2C+0.0000001152%29:
With x being negative, even powers of x will be positive but odd powers of x will be negative. This makes the 1st and 3rd arguments negative. This is invalid. So we must reject this "solution".

So the only solution to is x = x+=+root%2810%2C+0.0000001152%29.

P.S. Technically, the solution x+=+root%2810%2C+0.0000001152%29 should be rationalized. (A decimal is just a disguised fraction and radicals are not supposed to have fractions in them.) So the rest of this is rationalizing the solution:
x+=+root%2810%2C+0.0000001152%29
x+=+root%2810%2C+1152%2F10%5E10%29
x+=+root%2810%2C+1152%29%2Froot%2810%2C+10%5E10%29
x+=+root%2810%2C+1152%29%2F10