Question 817028: 2+log base3 ^(2x+5)- log base3^x = 4 , I can not figure out how to work it out properly. the answer is 5/7 Answer by jsmallt9(3758) (Show Source):
You can put this solution on YOUR website! First of all, there are no exponents in
so please do not use the exponentiation character, "^", unless there is an actual exponent in your expression. Your post should have been "2+log base 3 of (2x+5) - log base 3 of x = 4". Or you could teach yourself the algebra.com's syntax for entering formulas. Click on the "Show source" link above to see what I typed to get the logs to display so nicely.
Solving equations like this usually starts with using algebra and/or properties of logarithms to transform the equation into one of the following forms:
log(expression) = number
or
log(expression) = log(other_expression)
Without the "non-log" terms of 2 and 4 it would seem that the "all-log" second form will be more difficult to achieve. So we will aim for the first form. For this we will need to combine the two logs or somehow eliminate one of them and then isolate that logarithm. We can start by subtracting 2 from each side:
Now we combine the logs. They are not like terms so we cannot just subtract them. (Like logarithmic terms have the same bases and the same arguments. Our logs have the same bases, 3, but their arguments are different, 2x+5 and x.) But there is a property of logarithms, which we can use if the bases are the same and the coefficients are 1's. Out logs meet both requirements so:
And we now have the first form.
The second stage in these problems is to eliminate the logarithms. With the first form we eliminate the logarithm by rewriting the equation in exponential form. In general is equivalent to . Using this pattern on our equation we get:
which simplifies to:
Next we solve the equation (now that the logs are gone and the variable is exposed). Multiplying each side by x (to eliminate the fraction):
Subtracting 2x:
Dividing by 7:
Last we check. This is not optional! A check must be made to ensure that all bases and arguments of all logs are valid for the proposed solution. (Valid bases are positive but not 1 and valid arguments are positive.) Any "solution" that makes any base or any argument invalid must be rejected!
Use the original equation to check:
Checking x = 5/7:
We can probably already see that the bases are 3's (valid) and the arguments are both going to be positive (valid). So this solution checks out.
(