It should be "A"
On the number line, it shows that -4.5 is further and to the right of -2.3.
Hope that helps!
Move all the logarithms on the left hand side, and all the constants on the other:

Use the rule of logarithms

To rewrite the equation as

Evaluate 2 to the power of each side:

Multiply both sides by 2x:

Answer:
EG = 2 units
Step-by-step explanation:
Given that line q bisects EG at T , then
ET = TG ( substitute values )
x = x - 2 ( multiply through by 3 to clear the fraction )
x = 3x - 6 ( subtract x from both sides )
0 = 2x - 6 ( add 6 to both sides )
6 = 2x ( divide both sides by 2 )
3 = x
Then
ET =
x =
× 3 = 1
TG = x - 2 = 3 - 2 = 1
Thus
EG = ET + TG = 1 + 1 = 2 units
The easiest way is to make equations y=2400x+30000 and y=2000+36000
and then put that in the calculator and go to table and the point they intersect at is (15,66000)
You don't have the graph icon here, so we'll have to graph this parabola without it.
Your parabola is y = -x^2 + 3., which resembles y = a(x-h)^2 + k. We can tell immediately that this parabola opens down and that the vertex is (0,3).
Plot (0,3). Besides being the vertex, this point is also the max. of the function.
Now calculate four more points. Choose four arbitrary x-values, such as {-2, 1, 4, 5} and find the y value for each one. Plot the resulting four points. Draw a smooth curve thru them, remembering (again) that the vertex is at (0,3) and that the parabola opens down.