5. Additional Examples

Example 11

Solve for $x$ where $\mathrm{log}_{2}\mathrm{log} _{3} {3x}$ $=$ 0
($1$) $1$ ($2$) $4$ ($3$) $9$ ($4$) $16$

Solution

$\mathrm{log} _{2} \mathrm{log} _{3} {3x}$ $=$ $0$
⇒ $\mathrm{log} _{3} {3x}$ $=$ $1$
⇒ $3x$ $=$ $3$ ⇒ $x$ $=$ $1$

Answer: ($1$) $1$

Example 12

If $\mathrm{log} _{10} {0.04}$ $=$ $-1.398$, then $\mathrm{log} _{10} {400}$ $=$ ?
($1$)$1.602$ ($2$)$2.398$ ($3$) $2.602$ ($4$) $3.398$

Solution

$\mathrm{log} _{10} {400}$ $=$ $\mathrm{log} _{10} {(0.04 \times 10000)}$
$=$ $\mathrm{log} _{10} {0.04} + \mathrm{log} _{10} {10^4}$
$=$ $-1.398 + 4$ $=$ $2.602$

Answer: ($3$) $2.602$

Example 13

If $ x = \mathrm{log} _{10} {2}$ and $ y = \mathrm{log} _{10} {5}$ , then $\mathrm{log} _{5} {2}$ $=$ ?
($1$) $\dfrac{2y - x}{2y +x}$ ($2$)$\dfrac{2x + y}{2x - y}$ ($3$) $\dfrac{2 - x - 2y}{1 - x}$ ($4$) $\dfrac{2 - 2x - y}{1 - y}$

Solution

By substituting in each of the options we find that only option (3) satisfies

$= \dfrac{2 - x - 2y}{1 - x}$

$= \dfrac{2 - (\mathrm{log} \ 2 + 2\mathrm{log} \ 5)}{1 - \mathrm{log} \ 2}$

$= \dfrac{2 - \mathrm{log} (2 \times 5^{2})}{1 - \mathrm{log} \ 2}$ $= \dfrac{\mathrm{log} \ 100 - \mathrm{log} \ 50}{\mathrm{log} \ 10 - \mathrm{log} \ 2}$ $=\dfrac{\mathrm{log} \ 2}{\mathrm{log} \ 5} = \mathrm{log}_{5} {2}$

Answer: ($3$) $\dfrac{2 - x - 2y}{1 - x}$

Example 14

If $x$ $=$ $\mathrm{log} _{9} {144}$ and $y$ $=$ $\mathrm{log} _{12} {3}$, then $x$ $=$ ?
($1$) $\dfrac{1}{y}$ ($2$) $\dfrac{3}{2y}$ ($3$) $\dfrac{2y}{3}$ ($4$) $2y$

Solution

$x$ $=$ $\mathrm{log} _{3^2} {12^2}$ $=$ $\dfrac{2}{2} \mathrm{log} _{3} {12}$ $=$ $\dfrac{1}{\mathrm{log} _{12} {3}}$ $=$ $\dfrac{1}{y}$

Answer: ($1$) $\dfrac{1}{y}$

Example 15

$\mathrm{log} {x} + \mathrm{log} {(2x + 6)} $ $=$ $\mathrm{log} {(6 - 5x)}$. Solve for $x$.
($1$) $\dfrac{1}{2}$ ($2$) $-6$ ($3$) $\dfrac{1}{2}$ or $-6$ ($4$) None of the above

Solution

$\mathrm{log} {(2x^2 + 6x)}$ $=$ $\mathrm{log} {(6 - 5x)}$
⇒ $ 2x^2 + 11x - 6 $ $=$ $0$ ⇒ $-6$ or $\dfrac{1}{2}$

Log cannot be applied on a -ve number. At $x$ $=$ $-$ $ 6$, log $x$ will be negative and is, hence, rejected.

Answer: ($1$) $\dfrac{1}{2}$

Example 16

If $ \mathrm{log} {2}, \mathrm{log} {(2^{2x} + 3)}$ and $\mathrm{log} (2^{2x} + 27)$ are in arithmetic progression, then $x$ $=$ ?
($1$) $\mathrm{log} _{2} {3}$ ($2$) $2 \mathrm{log} _{2} {3}$ ($3$) $\mathrm{log} _{2} {\sqrt{5}}$ ($4$) $2 \mathrm{log} _{2} {5}$

Solution

Let $2^{2x}$ $=$ $y$
⇒ $ 2 \mathrm{log} {(y+3)}$ $=$ $\mathrm{log} {[2 \times (y+27)]}$
⇒ $\mathrm{log} {(y^2 + 6y +9 )}$ $=$ $\mathrm{log} {(2y + 54)}$
⇒ $y^2 + 4y - 45 $ $=$ $0$
⇒ $y$ $=$ $5$ or $-9$ [$-9$ is rejected.]

Where $x = 5$, $y$ $=$ $2^{2x}$ $=$ $2^{2 \mathrm{log} _{2} {\sqrt{2}}}$ $=$ $2^{\mathrm{log} _{2} {5}}$ $=$ $5$

Answer: ($3$) $\mathrm{log} _{2} {\sqrt{5}}$

Example 17

How many values of $x$ satisfy $\mathrm{log}_{x^2 - 2x +1}[\mathrm{log}_{x^2+x +1} {(3x^2 - 6x +7)}] = 0$
(1) 0 (2) 1 (3) 2 (4) More than 2

Solution

$\mathrm{log}_{x^2 - 2x +1}[\mathrm{log}_{x^2+x +1} {(3x^2 - 6x +7)}] = 0$

Removing the first logarithm,
⇒ $[\mathrm{log}_{x^2+x +1} {(3x^2 - 6x +7)}] = (x^2 - 2x +1)^0$

⇒$[log_{x^2+x+1} {(3x^2 - 6x +7)}] = 1$

Removing the next logarithm,
⇒ $(3x^2 - 6x +7) = (x^2 + x+1)^1$

⇒ $2x^2 - 7x +6 = 0$

Solving the quadratic equation, we get
⇒ $x$ = $\dfrac{3}{2}$, 2

It seems as though 2 solutions exist. However, there are variables in the bases and the value. We need to substitute these solutions and check if log holds good, i.e., $\mathrm{log}_{b} {a}$ is defined where $a$ > 0, $b$ > 0 and $b \neq 1$

When $x = 2$ is substituted into the base $x^2 - 2x +1$,

⇒ $x^2 - 2x +1 = 4 -4 +1 = 1$

As the base is not allowed to be 1, the solution $x = 2$ is rejected.

Substituting $x = \dfrac{3}{2} = 1.5$ into the following expressions we get

$x^2 - 2x +1 = 2.25 - 3 +1 = 0.25 $

$x^2 + x + 1 = 2.25 + 1.5 + 1 = 4.75$

(The bases are positive and not equal to 1)

$3x^2 - 6x + 7 = 6.75 - 9 + 7 = 4.75$

(The value is also positive)

Therefore, there is only 1 solution that exists, i.e., $x = 1.5$

Answer: (2) 1