(i) It is given that x satisfies the logarithm equation _a x=2 [ _a k- _a 2 ],where k>0, a>0, a 1. (a) Find x in terms of k, giving the answer in the form not involving logarithm. Suppose instead that x satisfies _x(5 y+1)=4+ _x 3 where, x>0, x 1, and y>0, y 1. (b) Solve the above equation expressing y in terms of x, giving the answer in a form not involving logarithm. (ii) Solve the equation 1 6 = (1 2 )^x and give your answer as single logarithm of base 2 .

(i) (a) _a x=2 ( _a k- _a 2 ) ll & _a x=2 _a k-2 _a 2 & _a x= _a k^2- _a 2^2 [Applying rule _a m^n=n _a m ] _a(x)= _a k^2 4 [Applying rule _a (m n )= _a m- _a n ] x=k^2 4 [Dropping log from both sides] l (b) _x(5 y+1)=4+ _x 3 _x(5 y+1)=4 _x x+ _x 3 [ _a a=1 ] _x(5 y+1)= _x(x)^4+ _x 3 [Applying rule _a m^n=n _a m ] _x(5 y+1)= _x (x^4 3 ) [Applying rule _a m^n=n _a m ] 5 y+1=3 x^4 [Dropping log from both sides] y=1 5 (3 x^4-1 ) . (ii) Given equation, l (1 6 )= (1 2 )^x (1 6 )= (2^ -1 )^x 6^ -1 =2^ -x _2 6^ -1 = _2 2^ -x [Taking with base 2 on both sides] - _2 6=-x _2 2 [Applying rule, _a m^n=n _a m ] - _2 6=-x [ _a=1 ] x= _2 6 x= _2(2 3) x= _2 2+ _2 3 [Applying rule . _a(m n)= _a m+ _a n ] x=1+ _2 3 [ _a a=1 ]

(i) It is given that x satisfies the logarithm equation _a x=2 [ …

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Question

(i) It is given that $x$ satisfies the logarithm equation $\log _a x=2\left[\log _a k-\log _a 2\right]$,where $k>0, a>0, a \neq 1$.
(a) Find $x$ in terms of $k$, giving the answer in the form not involving logarithm.
Suppose instead that $x$ satisfies
$\log _x(5 y+1)=4+\log _x 3$
where, $x>0, x \neq 1$, and $y>0, y \neq 1$.
(b) Solve the above equation expressing $y$ in terms of $x$, giving the answer in a form not involving logarithm.
(ii) Solve the equation $\frac{1}{6}=\left(\frac{1}{2}\right)^x$ and give your answer as single logarithm of base 2 .

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Answer

(i) (a) $
\log _a x=2\left(\log _a k-\log _a 2\right)
$
$
\begin{array}{ll}
\Rightarrow & \log _a x=2 \log _a k-2 \log _a 2 \\
\Rightarrow & \log _a x=\log _a k^2-\log _a 2^2
\end{array}
$
[Applying rule $\log _a m^n=n \log _a m$ ]
$
\Rightarrow \quad \log _a(x)=\log _a \frac{k^2}{4}
$
[Applying rule $\log _a\left(\frac{m}{n}\right)=\log _a m-\log _a n$ ]
$
\Rightarrow \quad x=\frac{k^2}{4}
$
[Dropping log from both sides]

$
\begin{array}{l}
\text { (b) }\log _x(5 y+1)=4+\log _x 3 \\
\log _x(5 y+1)=4 \log _x x+\log _x 3 \ {\left[\because \log _a a=1\right]} \\
\Rightarrow \quad \log _x(5 y+1)=\log _x(x)^4+\log _x 3 \\
\text { [Applying rule } \log _a m^n=n \log _a m \text { ] } \\
\Rightarrow \quad \log _x(5 y+1)=\log _x\left(x^4 \times 3\right) \\
\text { [Applying rule } \log _a m^n=n \log _a m \text { ] } \\
\Rightarrow5 y+1=3 x^4 \\
\text { [Dropping log from both sides] } \\
\Rightarrow \quad y=\frac{1}{5}\left(3 x^4-1\right) \text {. }
\end{array}
$

(ii) Given equation,
$
\begin{array}{l}
\Rightarrow \quad\left(\frac{1}{6}\right)=\left(\frac{1}{2}\right)^x \\
\Rightarrow \quad\left(\frac{1}{6}\right)=\left(2^{-1}\right)^x \\
\Rightarrow \quad 6^{-1}=2^{-x} \\
\Rightarrow \quad \log _2 6^{-1}=\log _2 2^{-x} \\
\text { [Taking } \log \text { with base } 2 \text { on both sides] } \\
\Rightarrow \quad-\log _2 6=-x \log _2 2 \\
\text { [Applying rule, } \log _a m^n=n \log _a m \text { ] } \\
\Rightarrow \quad-\log _2 6=-x \quad\left[\because \log _a=1\right] \\
\Rightarrow \quad\quad\quad\quad x=\log _2 6 \\
\Rightarrow \quad\quad\quad\quad x=\log _2(2 \times 3) \\
\Rightarrow \quad\quad\quad\quad x=\log _2 2+\log _2 3 \\
\text { [Applying rule } \left.\log _a(m n)=\log _a m+\log _a n\right] \\
\Rightarrow \quad\quad\quad\quad x=1+\log _2 3\left[\because \log _a a=1\right]
\end{array}
$