Arithmetic problem with logarithm

Question

This is the question which I am referring to

if log(base 3 of 2), log (base 3 of (2^(x)-5)), log(base 3 of (2^(x)-(7/2)) are in AP, find the value of x

My try and answer

I used the equation 2b= a+c , where a, b, c are in AP and solved for x so I got the answer log (base 2 of (6+(2)^(1/2)) and log (base 2 of (6-(2)^(1/2)) which comes out to be 2.89 and 2.17 respectively Is my way to solve question correct? Correct answer is 3.

Your use of parentheses is pretty confusing. First, in the last term, is it 2^x - (7/2) or (2^x-7)/2 ?? Also, you coul visit here to properly write mathematics in this site:http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-qu%E2%80%8C%E2%80%8Bick-reference — Timbuc, Jun 24 '15 at 16:35
No, you didn't in fact. At least, not what I remarked above! — Timbuc, Jun 24 '15 at 16:39

score 2 · Accepted Answer · answered Jun 24 '15 at 16:41

2

Read, understand and justify the following (assuming we have $\;2^x-\frac72\;$ in the last term):

$$\log_32\;,\;\;\log_3(2^x-5)=\log_32+d\;,\;\;\log_3\left(2^x-\frac72\right)=\log_32+2d$$

and also (we drop the subindex $\;3\;$ )

$$2\log(2^x-5)=\log2+\log\left(2^x-\frac72\right)=\log(2^{x+1}-7)\implies$$

$$2^{2x}-10\cdot2^x+25=2\cdot2^x-7$$

Substitute accordingly and get the quadratic $\;t^2-12t+32=(t-8)(t-4)\;$ , and end the exercise.

answered Jun 24 '15 at 16:41

Timbuc

34,191

I did it till that second last part and did blunder by using 18 instead of 32. – Kartik Watwani Jun 24 '15 at 16:45
@KartikWatwani Yes, sometimes our mistakes are littly, silly things...but which can affect greatly. Good it was only that. – Timbuc Jun 24 '15 at 16:53

Arithmetic problem with logarithm

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