Question

Log2(3x-3)=4 .................

Answer 1

Найдём одз:

3x-3>0

3x>3

x>1

$3x-3=2^4$

$3x-3=16$

$3x=19$

$x=\frac{19}{3}=6\frac{1}{3}$

Answer 2

log2(3x-3)=4 => x>1

3x-3=2^4 =>

3x = 19 =>

x=19/3 или x=6 целых 1/3

Answer 3

To solve the equation log2(3x-3) = 4, we need to isolate the variable x. The logarithmic equation is in the form of log(base) x = y, which means the logarithm with a given base of x equals y.

In this case, the base is 2, and the logarithm is 4:

log2(3x-3) = 4

To remove the logarithm, we can rewrite the equation in exponential form:

2^4 = 3x - 3

Now, simplify the left side:

16 = 3x - 3

Next, isolate the variable x by moving the constant term to the other side:

3x = 16 + 3 3x = 19

Finally, solve for x by dividing both sides by 3:

x = 19/3

So the solution to the equation is:

x = 19/3 or approximately 6.33.