Hi Dan
The way you solved the equation it looks like it should be
With the logs being base 3 and the x in the first log not being raised to the power. Is this what you meant?
There is this question
Solve for x in the following exponential an logarithmic equations
The solution is as follows
First use Product Rule to get
Then use Inversion Rules to get
Thus
And from there I can get the x values
While I understand the product rule I am not sure what is happening when using the inverse rule. I am unsure of where the "log" sign goes or where the number 3 comes from. Any sort of advice as to how you get these equations would be very helpful. Thank you
Hi Dan
The way you solved the equation it looks like it should be
With the logs being base 3 and the x in the first log not being raised to the power. Is this what you meant?
The basic rule concerning logs and exponentials is
$log_a(b) = c \iff a^c = b.$ Everything else follows from that.
The log function (not sign) and exponential function are inverses. You can go back and forth between them with this simple little conversion. You need to memorize it. It is definitional.
ASSUMING your problem is $log_3(x) + log_3(2x + 1) = 1$
THEN $log_3(x) + log_3(2x + 1) = 1 \implies log_3\{x(2x + 1)\} = 1.$ Product Rule.
SO $log_3(2x^2 + x) = 1.$ Simplification.
Now apply the basic inversion rule where $a = 3,\ b = 2x^2 + x,\ and\ c = 1.$
$log_3(2x^2 + x) = 1 \implies 2x^2 + x = 3^1 = 3 \implies 2x^2 + x - 3 = 0 \implies$
$(2x + 3)(x - 1) = 0 \implies x = -\ \dfrac{3}{2}\ or\ x = 1.$
One of those answers is wrong. Which one and why?