does someone know how to salve this ? x^(1+lg(x))=(10x)^-2
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Using guess and check on powers of 10 you'll find x = 1/10 will do the job.
Originally Posted by paulaa does someone know how to salve this ? x^(1+lg(x))=(10x)^-2 $\displaystyle x^{(1+log(x))}=(10x)^{-2}$ $\displaystyle logx^{(1+log(x))}=log(10x)^{-2}$ $\displaystyle logx(1+logx)=-2(1+logx)$ . . .
Originally Posted by Also sprach Zarathustra $\displaystyle x^{(1+log(x))}=(10x)^{-2}$ $\displaystyle logx^{(1+log(x))}=log(10x)^{-2}$ $\displaystyle logx(1+logx)=-2(1+logx)$ . . . $\displaystyle \text{lg}$ is ambiguous it can denote a base 10 or base 2 logarithm CB
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