Question

solve (3+log343)/(2+1/2log(49/4)+1/3log(1/125)​

Answer 1

Answer:

$3$

Step-by-step explanation:

$\dfrac{3+\log _{10}\left(343\right)}{2+\dfrac{1}{2}\log_ {10}\left(\dfrac{49}{4}\right)+\dfrac{1}{3}\log _{10}\left(\dfrac{1}{125}\right)}$

$\dfrac{1}{3}\log_ {10}\left(\dfrac{1}{125}\right)$

$\dfrac{1}{3}\log _{10}\left(\dfrac{1}{125}\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _a\left(\dfrac{1}{x}\right)=-\log _a\left(x\right)$

$=-\log_ {10}\left(125\right)$

$\mathrm{Rewrite\:}125\mathrm{\:in\:power-base\:form:}\quad 125=5^3$

$=-\log _{10}\left(5^3\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$

$\log_ {10}\left(5^3\right)=3\log _{10}\left(5\right)$

$=-3\log_ {10}\left(5\right)$

$=\dfrac{1}{3}\left(-3\log _{10}\left(5\right)\right)$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$=-\dfrac{1}{3}\cdot \:3\log_ {10}\left(5\right)$

$\mathrm{Multiply\:fractions}:\quad \:a\cdot \dfrac{b}{c}=\dfrac{a\:\cdot \:b}{c}$

$=-\dfrac{1\cdot \:3}{3}\log _{10}\left(5\right)$

$\mathrm{Cancel\:the\:common\:factor:}\:3$

$=-\log_ {10}\left(5\right)\cdot \:1$

$\mathrm{Multiply:}\:\log _{10}\left(5\right)\cdot \:1=\log_ {10}\left(5\right)$

$=-\log _{10}\left(5\right)$

$=\dfrac{3+\log_ {10}\left(343\right)}{2+\dfrac{1}{2}\log _{10}\left(\dfrac{49}{4}\right)-\log_ {10}\left(5\right)}$

$\log _{10}\left(343\right)$

$\log_ {10}\left(343\right)$

$\mathrm{Rewrite\:}343\mathrm{\:in\:power-base\:form:}\quad 343=7^3$

$=\log _{10}\left(7^3\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$

$\log_ {10}\left(7^3\right)=3\log _{10}\left(7\right)$

$=3\log_ {10}\left(7\right)$

$=\dfrac{3+3\log _{10}\left(7\right)}{2+\dfrac{1}{2}\log_ {10}\left(\dfrac{49}{4}\right)-\log _{10}\left(5\right)}$

$\dfrac{3+3\log_ {10}\left(7\right)}{2+\dfrac{1}{2}\log _{10}\left(\dfrac{49}{4}\right)-\log_ {10}\left(5\right)}$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$\dfrac{1}{2}\log_ {10}\left(\dfrac{49}{4}\right)=\log _{10}\left(\left(\dfrac{49}{4}\right)^{\dfrac{1}{2}}\right)$

$=\dfrac{3+3\log_ {10}\left(7\right)}{2+\log _{10}\left(\sqrt{\dfrac{49}{4}}\right)-\log_ {10}\left(5\right)}$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\dfrac{a}{b}\right)$

$\log_ {10}\left(\left(\dfrac{49}{4}\right)^{\dfrac{1}{2}}\right)-\log _{10}\left(5\right)=\log_ {10}\left(\dfrac{\sqrt{\dfrac{49}{4}}}{5}\right)$

$=\dfrac{3+3\log _{10}\left(7\right)}{2+\log_ {10}\left(\dfrac{\sqrt{\dfrac{49}{4}}}{5}\right)}$

$\dfrac{\left(\dfrac{49}{4}\right)^{\dfrac{1}{2}}}{5}$

$\dfrac{\sqrt{\dfrac{49}{4}}}{5}$

$\sqrt{\dfrac{49}{4}}$

$\mathrm{Apply\:radical\:rule\:}\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0$

$=\dfrac{\sqrt{49}}{\sqrt{4}}$

$=\dfrac{7}{2}$

$=\dfrac{\dfrac{7}{2}}{5}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \dfrac{\dfrac{b}{c}}{a}=\dfrac{b}{c\:\cdot \:a}$

$=\dfrac{7}{2\cdot \:5}$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:5=10$

$=\dfrac{7}{10}$

$=\dfrac{3+3\log _{10}\left(7\right)}{2+\log_ {10}\left(\dfrac{7}{10}\right)}$

$\dfrac{3+3\log _{10}\left(7\right)}{2+\log_ {10}\left(\dfrac{7}{10}\right)}$

$\log _{10}\left(\dfrac{7}{10}\right)$

$\log_ {10}\left(\dfrac{7}{10}\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(\dfrac{a}{b}\right)=\log _c\left(a\right)-\log _c\left(b\right)$

$\log_ {10}\left(\dfrac{7}{10}\right)=\log _{10}\left(7\right)-\log_ {10}\left(10\right)$

$=\log _{10}\left(7\right)-\log_ {10}\left(10\right)$

$\mathrm{Simplify}\:\log _{10}\left(10\right)$

$\log_ {10}\left(10\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _a\left(a\right)=1$

$=1$

$=\log_ {10}\left(7\right)-1$

$=\dfrac{3+3\log _{10}\left(7\right)}{2+\log_ {10}\left(7\right)-1}$

$\mathrm{Subtract\:the\:numbers:}\:2-1=1$

$=\dfrac{3+3\log _{10}\left(7\right)}{\log_ {10}\left(7\right)+1}$

$\dfrac{3+3\log _{10}\left(7\right)}{\log_ {10}\left(7\right)+1}$

$\mathrm{Refine\:to\:a\:decimal\:form}$

$=3$

Answer 2

Answer:

3

Step-by-step explanation:

\dfrac{3+\log _{10}\left(343\right)}{2+\dfrac{1}{2}\log_ {10}\left(\dfrac{49}{4}\right)+\dfrac{1}{3}\log _{10}\left(\dfrac{1}{125}\right)}

\dfrac{1}{3}\log_ {10}\left(\dfrac{1}{125}\right)

\dfrac{1}{3}\log _{10}\left(\dfrac{1}{125}\right)

\mathrm{Apply\:log\:rule}:\quad \log _a\left(\dfrac{1}{x}\right)=-\log _a\left(x\right)

=-\log_ {10}\left(125\right)

\mathrm{Rewrite\:}125\mathrm{\:in\:power-base\:form:}\quad 125=5^3

=-\log _{10}\left(5^3\right)

\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)

\log_ {10}\left(5^3\right)=3\log _{10}\left(5\right)

=-3\log_ {10}\left(5\right)

=\dfrac{1}{3}\left(-3\log _{10}\left(5\right)\right)

\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a

=-\dfrac{1}{3}\cdot \:3\log_ {10}\left(5\right)

\mathrm{Multiply\:fractions}:\quad \:a\cdot \dfrac{b}{c}=\dfrac{a\:\cdot \:b}{c}

=-\dfrac{1\cdot \:3}{3}\log _{10}\left(5\right)

\mathrm{Cancel\:the\:common\:factor:}\:3

=-\log_ {10}\left(5\right)\cdot \:1

\mathrm{Multiply:}\:\log _{10}\left(5\right)\cdot \:1=\log_ {10}\left(5\right)

=-\log _{10}\left(5\right)

=\dfrac{3+\log_ {10}\left(343\right)}{2+\dfrac{1}{2}\log _{10}\left(\dfrac{49}{4}\right)-\log_ {10}\left(5\right)}

\log _{10}\left(343\right)

\log_ {10}\left(343\right)

\mathrm{Rewrite\:}343\mathrm{\:in\:power-base\:form:}\quad 343=7^3

=\log _{10}\left(7^3\right)

\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)

\log_ {10}\left(7^3\right)=3\log _{10}\left(7\right)

=3\log_ {10}\left(7\right)

=\dfrac{3+3\log _{10}\left(7\right)}{2+\dfrac{1}{2}\log_ {10}\left(\dfrac{49}{4}\right)-\log _{10}\left(5\right)}

\dfrac{3+3\log_ {10}\left(7\right)}{2+\dfrac{1}{2}\log _{10}\left(\dfrac{49}{4}\right)-\log_ {10}\left(5\right)}

\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)

\dfrac{1}{2}\log_ {10}\left(\dfrac{49}{4}\right)=\log _{10}\left(\left(\dfrac{49}{4}\right)^{\dfrac{1}{2}}\right)

=\dfrac{3+3\log_ {10}\left(7\right)}{2+\log _{10}\left(\sqrt{\dfrac{49}{4}}\right)-\log_ {10}\left(5\right)}

\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\dfrac{a}{b}\right)

\log_ {10}\left(\left(\dfrac{49}{4}\right)^{\dfrac{1}{2}}\right)-\log _{10}\left(5\right)=\log_ {10}\left(\dfrac{\sqrt{\dfrac{49}{4}}}{5}\right)

=\dfrac{3+3\log _{10}\left(7\right)}{2+\log_ {10}\left(\dfrac{\sqrt{\dfrac{49}{4}}}{5}\right)}

\dfrac{\left(\dfrac{49}{4}\right)^{\dfrac{1}{2}}}{5}

\dfrac{\sqrt{\dfrac{49}{4}}}{5}

\sqrt{\dfrac{49}{4}}

\mathrm{Apply\:radical\:rule\:}\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0

=\dfrac{\sqrt{49}}{\sqrt{4}}

=\dfrac{7}{2}

=\dfrac{\dfrac{7}{2}}{5}

\mathrm{Apply\:the\:fraction\:rule}:\quad \dfrac{\dfrac{b}{c}}{a}=\dfrac{b}{c\:\cdot \:a}

=\dfrac{7}{2\cdot \:5}

\mathrm{Multiply\:the\:numbers:}\:2\cdot \:5=10

=\dfrac{7}{10}

=\dfrac{3+3\log _{10}\left(7\right)}{2+\log_ {10}\left(\dfrac{7}{10}\right)}

\dfrac{3+3\log _{10}\left(7\right)}{2+\log_ {10}\left(\dfrac{7}{10}\right)}

\log _{10}\left(\dfrac{7}{10}\right)

\log_ {10}\left(\dfrac{7}{10}\right)

\mathrm{Apply\:log\:rule}:\quad \log _c\left(\dfrac{a}{b}\right)=\log _c\left(a\right)-\log _c\left(b\right)

\log_ {10}\left(\dfrac{7}{10}\right)=\log _{10}\left(7\right)-\log_ {10}\left(10\right)

=\log _{10}\left(7\right)-\log_ {10}\left(10\right)

\mathrm{Simplify}\:\log _{10}\left(10\right)

\log_ {10}\left(10\right)

\mathrm{Apply\:log\:rule}:\quad \log _a\left(a\right)=1

=1

=\log_ {10}\left(7\right)-1

=\dfrac{3+3\log _{10}\left(7\right)}{2+\log_ {10}\left(7\right)-1}

\mathrm{Subtract\:the\:numbers:}\:2-1=1

=\dfrac{3+3\log _{10}\left(7\right)}{\log_ {10}\left(7\right)+1}

\dfrac{3+3\log _{10}\left(7\right)}{\log_ {10}\left(7\right)+1}

\mathrm{Refine\:to\:a\:decimal\:form}

=3