log_3(2x - 1) = 2, Find x.
Here's what I've done:
log_3(2x) * log_3(1) = 2
log2x/log3 * log1/log3 = 2
trial and error...
log2 (2*1.4)/log3 * 0 = 2.03...
x = 1.4
I think I did it wrong because anything that multiplies with 0 gives you 0...
3^log_3(2x - 1) = 3^2
but 3^log3 a = a for any a
so
2 x - 1 = 3^2 = 9
2 x = 10
x = 5
thank you! so if I get a problem like that, all I do is raise the base to the power of the log?
yep
To solve the equation log₃(2x - 1) = 2, you can follow these steps:
Step 1: Express the equation in exponential form.
Using the definition of logarithms, rewrite the equation log₃(2x - 1) = 2 as 3² = 2x - 1.
Step 2: Solve for x.
Simplify the equation: 9 = 2x - 1.
Add 1 to both sides to isolate the term with x: 9 + 1 = 2x.
Simplify further: 10 = 2x.
Divide both sides by 2: 10/2 = x.
Simplify: x = 5.
Therefore, the solution to the equation log₃(2x - 1) = 2 is x = 5.