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$\log_2 X$, $\log_2 (X+9)$ 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.

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2  
to begin with: do you mean $log_{2}X, log_{2}(x+9), log_{2}(x+45)$ or $log(2^x), log(2^{x+9}), log(2^{x+45})$? clarify it please so that the problem can be solved. –  Nikhil Bellarykar Jan 5 '12 at 7:19
    
I’ve edited it into the form that you actually wrote; is that what you intended, or did you want $\log_2 X$, $\log_2 (X+9)$, and $\log_2 (X+45)$? –  Brian M. Scott Jan 5 '12 at 7:27
1  
Never mind: the problem makes no sense unless I reinterpret it with logs base $2$, so I’ve changed the edit. –  Brian M. Scott Jan 5 '12 at 7:33
    
@BrianM.Scott. I think this will not work since you get $X\log 2-(X+9)\log 2=(X+9)\log 2-(X+45)\log 2$ which is false as you can check. Maybe he means the second one. –  smanoos Jan 5 '12 at 7:34
    
Must be second, $X=3$. –  André Nicolas Jan 5 '12 at 7:40

3 Answers 3

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.

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We get the false result $-9\log2=-36\log2$. –  Nikhil Bellarykar Jan 5 '12 at 7:32
    
@NikhilBellarykar. The question is fixed now. –  smanoos Jan 5 '12 at 7:42
    
yeah now it is. –  Nikhil Bellarykar Jan 5 '12 at 8:20

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.

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  1. $X=3$

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

  3. 5th term $\log_2(768)$

I leave the way to you...

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