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If

  • ${\mathrm{\log_{10}x}}^{\mathrm{}_{}}$ = a

and

  • ${\mathrm{\log_{10}y}}^{\mathrm{}_{}}$ = c

Express: ${\mathrm{Log}10}^{\mathrm{}_{}}($$\frac{\mathrm{100x^3 * y^-1/2}}{\mathrm{y^2}_{}})$$ $ in terms of a and c.

= ${\mathrm{\log_{10}}}^{\mathrm{}_{}}($$\frac{\mathrm{100x^3 * 1/√y}}{\mathrm{y^2}_{}})$$ $

= ${\mathrm{\log_{10}}}(^{\mathrm{}_{}}$$\frac{\mathrm{100x^3}}{\mathrm{√y}_{}}* $$\frac{\mathrm{1}}{\mathrm{y^2}_{}} )$$ $

= ${\mathrm{\log_{10} }}^{\mathrm{}_{}}($$\frac{\mathrm{100x^3}}{\mathrm{y^(1/2) *y^2 }_{}})$$ $

= ${\mathrm{\log_{10} }}^{\mathrm{}_{}}($$\frac{\mathrm{100x^3}}{\mathrm{y^(5/2) }_{}})$$ $

= ${\mathrm{\log_{10}}}{\mathrm{100x^3}_{}} $ - ${\mathrm{\log_{10} }}{\mathrm{y^(5/2)}_{}} $

Where do I go from here? I am bad at editing so sorry if its too small.

  • 1
    Your $10$ is the base? – 2017-02-14
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    @tommyxu3 yes. I am sorry i dont know how to edit that... – 2017-02-14
  • 0
    @Utsav: \log_{10}x to get $\log_{10}x$ – 2017-02-14
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    Also use \sqrt{y} to obtain $\sqrt{y}$ – 2017-02-14

3 Answers 3

1

If your problem is $\log_{10}\frac{100x^3}{y^{\frac{5}{2}}} = \log_{10}100x^3 - \log_{10}y^{\frac{5}{2}},$ then you can proceed as $$\log_{10}100x^3 - \log_{10}y^{\frac{5}{2}} = \log_{10}100+\log_{10}x^3-\log_{10}y^{\frac{5}{2}} = 2+3a-\frac{5}{2}c$$ with $\log u^v = v\log u.$

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$\log_{10}{\frac{100x^3y^{-\frac{1}{2}}}{y^3}}=\log_{10}^{\frac{100x^3}{y^\frac{5}{2}}}=\log_{10}100+\log_{10}{x^3}-\log_{10}{y^\frac{5}{2}}=2+3\log_{10}x-\frac{5}{2}\log_{10}y=\boxed{2+3a-\frac{5}{2}c}$

0

The products become sums, the quotients, subtractions and the exponents, factors.

$$\log ab=\log a+\log b,\ \log\frac ab=\log a-\log b,\ \log a^b=b\log a.$$

Hence

$$\log_{10}100+3\log_{10}x-\frac12\log_{10}y-2\log_{10}y,\\=2+3a-\frac52c.$$