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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 $x+3$.2012-04-13
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    $$\log 2^{x-1}-\log 2=\log 2^{x+3}-\log 2^{x-1}$$2012-04-13

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