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Question Number 113800 by Aina Samuel Temidayo last updated on 15/Sep/20

$$\log \left(\mathrm{ab}\right)-\log \mid \mathrm{b}\mid =$$$$$$$$\mathrm{A}.\log \left(\mathrm{a}\right)\mathrm{B}.\log \mid \mathrm{a}\mid \mathrm{C}.-\log \left(\mathrm{a}\right)\mathrm{D}.$$$$\mathrm{None}\mathrm{of}\mathrm{these}.$$

Answered by MJS_new last updated on 15/Sep/20

$$\left(1\right)a<0\wedge b<0$$$$\log ab-\log \mid b\mid =\log (-a)$$$$\left(2\right)a>0\wedge b>0$$$$\log ab-\log \mid b\mid =\log a$$$$\Rightarrow \mathrm{answer}\mathrm{is}B.\log \mid a\mid$$

Commented by MJS_new last updated on 15/Sep/20

$$\mathrm{test}\mathrm{it}.$$$$\mathrm{i}.\mathrm{e}.a=-3\wedge b=-5$$$$\log ab=\log 15$$$$\log \mid b\mid =\log 5$$$$\log 15-\log 5=\log 3=\log -a$$$$\Rightarrow \mathrm{if}a<0\wedge b<0\mathrm{we}\mathrm{get}\log -a(\mathrm{but}a<0\Rightarrow -a>0)$$$$\mathrm{if}a>0\wedge b>0\mathrm{we}\mathrm{get}\log a$$$$\Rightarrow \mathrm{we}\mathrm{get}\log \mid a\mid$$

Commented by Aina Samuel Temidayo last updated on 15/Sep/20

$$\log \left(\mathrm{ab}\right)=\log \left(\mathrm{a}\right)+\log \left(\mathrm{b}\right)$$$$\Rightarrow \mathrm{a},\mathrm{b}\mathrm{is}+\mathrm{ve}$$$$\Rightarrow \log \mid \mathrm{b}\mid =\log \mathrm{b}$$$$\Rightarrow \log \left(\mathrm{ab}\right)-\log \mid \mathrm{b}\mid$$$$=\log \left(\mathrm{ab}\right)-\log \left(\mathrm{b}\right)$$$$=\log \left(\frac{\mathrm{ab}}{\mathrm{b}}\right)=\log \left(\mathrm{a}\right).$$$$$$

Commented by Aina Samuel Temidayo last updated on 15/Sep/20

$$\mathrm{a},\mathrm{b}\mathrm{cannot}\mathrm{be}-\mathrm{ve}.$$

Commented by Aina Samuel Temidayo last updated on 15/Sep/20

$$\mathrm{From}\left(1\right),\mathrm{you}\mathrm{said}\mathrm{a}<0\wedge \mathrm{b}<0$$$$\Rightarrow \mathrm{ab}>0$$$$\mathrm{Note}\mathrm{that}\mid \mathrm{b}\mid >0$$$$\Rightarrow \frac{\mathrm{ab}}{\mid \mathrm{b}\mid }\mathrm{is}+\mathrm{ve}$$$$$$

Commented by MJS_new last updated on 15/Sep/20

$$\mathrm{you}\mathrm{misunderstand}\mid x\mid \mathrm{it}\mathrm{seems}.\mathrm{remember}$$$$\mid x\mid =\{\begin{array}{l}-x;x<0\\ x;x⩾0\end{array}$$

Commented by Aina Samuel Temidayo last updated on 15/Sep/20

$$\mathrm{since}\log \left(\mathrm{ab}\right)=\log \left(\mathrm{a}\right)+\log \left(\mathrm{b}\right)$$$$\mathrm{a},\mathrm{b}\mathrm{should}\mathrm{not}\mathrm{be}\mathrm{negative}..$$

Commented by MJS_new last updated on 15/Sep/20

$$\mathrm{no}.\log ab=\log a+\log b\mathrm{only}\mathrm{if}a>0\wedge b>0$$$$\log ab\mathrm{exists}\mathrm{if}ab>0\Rightarrow a>0\wedge b>0\vee a<0\wedge b<0$$$$\mathrm{if}a<0\wedge b<0\Rightarrow \log ab\ne \log a+\log b$$$$\mathrm{but}\mathrm{we}\mathrm{can}\mathrm{still}\mathrm{calculate}\log ab-\log \mid b\mid$$$$\mathrm{because}ab>0\wedge \mid b\mid >0\mathrm{and}\mathrm{we}\mathrm{get}\log \mid a\mid$$$$\mathrm{because}a<0\mathrm{but}\mathrm{the}\mathrm{argument}\mathrm{we}\mathrm{get}\mathrm{is}>0$$

Commented by Aina Samuel Temidayo last updated on 15/Sep/20

$$\mathrm{Ok}.\mathrm{Thanks}\stackrel{}{.}$$

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