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$\log 2,\log 2^{x-1}$, and $\log 2^{x+3}$ are $3$ consecutive terms of an arithmetic progression; find (i) the value of $x$;

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Please check that I correctly interpreted what you wrote when I made the exponents $x-1$ and $x+3$. – Brian M. Scott Apr 13 '12 at 5:19
$$\log 2^{x-1}-\log 2=\log 2^{x+3}-\log 2^{x-1}$$ – pedja Apr 13 '12 at 5:27

If $\log 2,\log 2^{x-1}$, and $\log 2^{x+3}$ are three consecutive terms of an arithmetic progression, there must be a common difference $d$ such that $\log 2^{x-1}=\log 2+d$ and $\log 2^{x+3}=\log 2^{x-1}+d=$ $\log 2+2d$. From basic properties of the logarithm we know that $$\log 2^{x-1}=(x-1)\log 2$$ and $$\log 2^{x+3}=(x+3)\log 2\;,$$ so $$\left\{\begin{align*}&(x-1)\log 2=\log 2+d\\&(x+3)\log 2=\log 2+2d\;.\end{align*}\right.$$ This system can easily be solved for $d$ and $x$.

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Using $\log(a^b)=b\log a$ and dividing by $\log2$ we see that $1,x-1,x+3$ must be an arithmetic progression. But it's clear that $x-5,x-1,x+3$ is an arithmetic progression. So $x\dots$

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