Question 821505: Kindly help me answer these items/how to answer them:
I. Solve the ff. equations. Answers with log should not contain coefficients
1)(1/36)^2-x = 216 ^ 6x-2
2) 2log of x to the base 5= log (base)5 (8x-15)
3)log (base) 2 (3 + log (base) 3 (x-3))=2
4)10^log (x^2-6) =3
5) 2^x (3^(x-2))=5^2x
6)5(25^x-1)= 1
7)2^ 2x- 2^(x+1)= -1
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! Please...- Don't include so many problems in a single post. Tutors are unlikely to help when so much work is to be done.
- Use parentheses around...
- Exponents
- Arguments to functions (like log, sin, cos, etc.)
- Radciands ("radicand" is the name for the expression inside of a radical)
- Numerators and denominators
- ...anything to make the problem clear and unambiguous.
... especially if they are not just a positive integer or variable.
On the other hand, good job on posting the logs! They are very difficult to post clearly.
I'm going to do just #3 (because it appears to be the hardest one):
To solve for x we need to eliminate the logs. Eliminating a log usually starts with using algebra and/or properties of logs to transform the equation into one of the following forms:
log(expression) = number
or
log(expression) = log(other_expression) [Note: The bases must match in this form!]
Your equation, fortunately, is already in the first form!. With the first form the next step is to rewrite the equation in exponential form. In general,  is equivalent to  . Using this pattern on your equation we get:
which simplifies to:
We still have a log to eliminate. So we are again looking to get the equation into one of the forms above. If we subtract 3 from each side we will have the first form again:
Again we rewrite this in exponential form:
which simplifies to:
Now that the logs are gone we can "get at" the x and solve for it. Adding 3 to each side:
x = 6
Finally we check. This is not optional when solving logarithmic equations like this one! A check must be made to ensure that all bases and arguments of all logs remain valid. (Valid bases are positive but not 1 and valid arguments are positive.) If a "solution" makes any base or any argument of any log invalid, then it must be rejected (even if it is the only "solution" you found).
Use the original equation to check:
Checking x = 6:
Simplifying...
By definition,  . Replacing this log with a 1:
At this point we should be able to see that the base is 2 and the argument will be 4 (after we add). Both the base and the argument are valid. So our solution passes the check.
P.S. If x = 6 had failed the check then we would have had to reject it. And since it was the only "solution" we came up with, rejecting it would mean that the equation had no solutions!
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