log 128 - log 18 / log 16 - log 6

Question

ScanSolve AI Tutor · Accepted Answer

Hapa kuna hatua za kurahisisha usemi wa logariti:

Usemi uliotolewa ni:

\frac{\log 128 - \log 18}{\log 16 - \log 6}

Tumia sheria ya logariti $\log A - \log B = \log \left(\frac{A}{B}\right)$ kwa nambari (numerator).
$\log 128 - \log 18 = \log \left(\frac{128}{18}\right)$
Rahisisha sehemu ndani ya logariti:
$\log \left(\frac{128}{18}\right) = \log \left(\frac{64}{9}\right)$
Tumia sheria ya logariti $\log A - \log B = \log \left(\frac{A}{B}\right)$ kwa kinyume (denominator).
$\log 16 - \log 6 = \log \left(\frac{16}{6}\right)$
Rahisisha sehemu ndani ya logariti:
$\log \left(\frac{16}{6}\right) = \log \left(\frac{8}{3}\right)$
Andika upya usemi mzima kwa kutumia matokeo ya hatua za awali.
$\frac{\log \left(\frac{64}{9}\right)}{\log \left(\frac{8}{3}\right)}$
Tambua uhusiano kati ya hoja (arguments) za logariti.
Kumbuka kuwa $\frac{64}{9}$ inaweza kuandikwa kama $\left(\frac{8}{3}\right)^2$ .
Hivyo, usemi unakuwa:
$\frac{\log \left(\left(\frac{8}{3}\right)^2\right)}{\log \left(\frac{8}{3}\right)}$
Tumia sheria ya logariti $\log A^n = n \log A$ kwa nambari.
$\log \left(\left(\frac{8}{3}\right)^2\right) = 2 \log \left(\frac{8}{3}\right)$
Badilisha hii kwenye usemi.
$\frac{2 \log \left(\frac{8}{3}\right)}{\log \left(\frac{8}{3}\right)}$
Futa neno la kawaida $\log \left(\frac{8}{3}\right)$ kutoka nambari na kinyume.

Jibu la mwisho ni:
$2$

Answer