arithmetic progression involving logarithm $\log_2 X$, $\log_2 (X+9)$ and $\log_2(X+45)$ are 3 consecutive terms of an arithmetic progression; find 
$\qquad$(i) the value of X;
$\qquad$(ii) the first term and the common difference; and
$\qquad$(iii) the 5th term as a single logarithm.
 A: Hint:
Solve the equation $$\log_2X-\log_2(X+9)=\log_2(X+9)-\log_2(X+45)$$ 
for X. This follows from the fact that the difference between any two consecutive terms is the same.
So you get
$$\frac{X}{X+9}=\frac{X+9}{X+45}$$ which gives $X=3$
Observe that the common difference will be $\log_2(X+9)-\log_2X$.
You can now obtain the 5th term once you know the common difference.
A: Since it’s an arithmetic progression, let $d$ be the constant difference; then $$\log_2(X+9)=\log_2 X+d\;,\tag{1}$$ and $$\log_2(X+45)=\log_2(X+9)+d\;.\tag{2}$$ Let $a=2^d$, so that $\log_2 a=d$. Then $(1)$ becomes $$\log_2(X+9)=\log_2 X+\log_2 a=\log_2 aX\;,$$ so $X+9=aX$, $X=aX-9$, and we can rewrite $(2)$ as $$\log_2(aX+36)=\log_2 aX+\log_2 a=\log_2 a^2X$$ to get $$aX+36=a^2X=a(X+9)=aX+9a\;.$$
Now you’re home free: just solve for $a$ to get $a=4$, which immediately lets you solve for $d$, $X$, and everything else.
A: *

*$X=3$

*first term $\log_23$ and difference $\Delta = 2$

*5th term $\log_2(768)$ 
I leave the way to you...
