Fill in the blanks.

Question 11 Mark

$\log \left(\frac{1}{81}\right)$ to the base 9 is equal to _________________

Answer

-2 , because
$
\begin{aligned}
\log _9\left(\frac{1}{81}\right) & =\log _9\left(\frac{1}{9^2}\right)=\log _9(9)^{-2} \\
& =-2 \log _9 9=-2 \times 1=-2
\end{aligned}
$
[Applying rules $\log _a(m)^n=n \log _a m$ and $\left.\log _a a=1\right]$

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Question 21 Mark

Given, $\log 2=0.3010$ and $\log 3=0.4771$, the value of $\log 6$ is _________________

Answer

0.7781 , because
$
\begin{aligned}
\log 6 & =\log (2 \times 3) \\
& =\log 2+\log 3 \\
& =0.3010+0.4771 \\
& =0.7781
\end{aligned}
$

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Question 31 Mark

$\log _{\sqrt{2}} 64$ is equal to _________________

Answer

12, because
$\begin{aligned} \log _{\sqrt{2}} 64 & =\log _{\sqrt{2}}(\sqrt{2})^{12} \\ & =12 \log _{\sqrt{2}}(\sqrt{2}) \\ & =12 \times 1 \\ & =12\end{aligned}$

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Question 41 Mark

The value of $\log 0.0001$ to the base 0.1 is _________________

Answer

4, because$
\begin{array}{r}
\log _{0.1} 0.0001=\log _{0.1}(0.1)^4=4 \log _{0.1}(0.1)=4 \times 1=4 \\
{\left[\text { Applying rules } \log _a(m)^n=n \log _a m \text { and } \log _a a=1\right]}
\end{array}
$

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Question 51 Mark

$\frac{\left(81 x^4\right)^{\frac{1}{4}}}{y^{-8}}$ is equal to _________________

Answer

$3 x y^8$, because
$
\frac{\left(81 x^4\right)^{1 / 4}}{y^{-8}}=\frac{\left[(3)^4 x^4\right]^{1 / 4}}{y^{-8}}
$
$\begin{array}{l}\qquad \frac{3 x}{y^{-8}}=3 x y^8 \\ {\left[\text { Using }\left(a^{m n}\right)^n=a^{m n} \text { and } \frac{1}{a^{-n}}=a^n\right]}\end{array}$

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Question 61 Mark

The value of $\left(\frac{2 p^2 q^3}{3 x y}\right)^{\circ}$, where $p, q, x, y \neq 0$ is equal to _________________

Answer

1, because According to zero index, any base raised to the power zero is equal to 1.

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Question 71 Mark

$\left\{\left(x^n\right)^{n-\frac{1}{n}}\right\}^{\frac{1}{n+1}}$ is equal to _________________

Answer

$x^{n-1}$, because

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Question 81 Mark

$x^{a-b} \times x^{b-c} \times x^{c-a}$ is equal to _________________

Answer

1, because
$\begin{aligned} x^{a-b} \times x^{b-c} \times x^{c-a}= & x^{a-b+b-c+c-a} \\ & \quad\left[ Using a^m \times a^n=a^{m+n}\right] \\ = & x^0=1 .\end{aligned}$

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