What is 4*log_{0.1}(3)-2/3*log_{0.1}(27)+2*log_{0.1}(6) ?

The solution to 4*log_{0.1}(3)-2/3*log_{0.1}(27)+2*log_{0.1}(6) is -2log_{10}(18)

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simplify 4log_{0.1}(3)-2/3log_{0.1}(27)+2*log_{0.1}(6)

simplify

4 \cdot lo g_{0.1} (3) - \frac{2}{3} \cdot lo g_{0.1} (27) + 2 \cdot lo g_{0.1} (6)

- 2 lo g_{10} (18)

Decimal Notation

- 2.51054 \dots

Solution steps

4 lo g_{0.1} (3) - \frac{2}{3} lo g_{0.1} (27) + 2 lo g_{0.1} (6)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{2}{3} lo g_{0.1} (27) = lo g_{0.1} (2 7^{\frac{2}{3}})

= 4 lo g_{0.1} (3) - lo g_{0.1} (2 7^{\frac{2}{3}}) + 2 lo g_{0.1} (6)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 lo g_{0.1} (6) = lo g_{0.1} (6^{2})

= 4 lo g_{0.1} (3) - lo g_{0.1} (2 7^{\frac{2}{3}}) + lo g_{0.1} (6^{2})

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

lo g_{0.1} (6^{2}) - lo g_{0.1} (2 7^{\frac{2}{3}}) = lo g_{0.1} (\frac{6 ^{2}}{2 7 ^{\frac{2}{3}}})

= 4 lo g_{0.1} (3) + lo g_{0.1} (\frac{6 ^{2}}{2 7 ^{\frac{2}{3}}})

\frac{6 ^{2}}{2 7 ^{\frac{2}{3}}} = 4

= 4 lo g_{0.1} (3) + lo g_{0.1} (4)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

4 lo g_{0.1} (3) = lo g_{0.1} (3^{4})

= lo g_{0.1} (3^{4}) + lo g_{0.1} (4)

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

lo g_{0.1} (3^{4}) + lo g_{0.1} (4) = lo g_{0.1} (3^{4} \cdot 4)

= lo g_{0.1} (3^{4} \cdot 4)

3^{4} \cdot 4 = 324

= lo g_{0.1} (324)

Convert

0.1

to a fraction:

\frac{1}{10}

= lo g_{\frac{1}{10}} (324)

Apply log rule:

lo g_{\frac{1}{a}} (x) = - lo g_{a} (x)

= - lo g_{10} (324)

Rewrite

324

in power-base form:

324 = 1 8^{2}

= - lo g_{10} (1 8^{2})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

lo g_{10} (1 8^{2}) = 2 lo g_{10} (18)

= - 2 lo g_{10} (18)

2 lo g_{2} (7) + lo g_{2} (3)

ln (\frac{1}{8}) - ln (\frac{1}{2})

ln (\frac{x}{7})

ln (\frac{0.4}{0.6})

ln (\frac{18038.519}{72327.395})

What is 4*log_{0.1}(3)-2/3*log_{0.1}(27)+2*log_{0.1}(6) ?
The solution to 4*log_{0.1}(3)-2/3*log_{0.1}(27)+2*log_{0.1}(6) is -2log_{10}(18)