Problem

$1og_{8}25$ $=$ $\dfrac {1og^{25}} {10g8}$ $=$ $\dfrac {1og^{5^{2}}} {10g^{2^{3}}}$ $\dfrac {2log5} {31og^{2}}$ $\left(^{.:}$ $a^{b}=$ $1o_{g}$ $bloga\right)$
$bne$ $10g_{8}25=\dfrac {210g^{5}} {310g2}=\dfrac {2\times 0.6989} {3\times 0.3010}$ $\left(10g^{5}=0.6989$ $\right)$
$0g^{2}$ $=0.3010$
$=\dfrac {1.3978} {0.9030}$
$=1.5479$ $2$ $1548$ $AM$ $\bar{1C} $

10th-13th grade

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Solution

Qanda teacher - AskShivani

what is your doubt in question?

Student

can't understand the 2 step

Qanda teacher - AskShivani

it is a log property

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