Logarithms — Chapter Test

Expand $\text{log}_a \space {\sqrt[3]{x^7y^8 ÷ \sqrt[4]{z}}}$

Answer

Given,

$\Rightarrow \text{log}_a \space {\sqrt[3]{x^7y^8 ÷ \sqrt[4]{z}}} \\[1em] \Rightarrow \text{log}_a \space {(x^7y^8 ÷ \sqrt[4]{z})^{\dfrac{1}{3}}} \\[1em] \Rightarrow \dfrac{1}{3}\text{log}_a \space {(x^7y^8 ÷ \sqrt[4]{z})} \\[1em] \Rightarrow \dfrac{1}{3}[\text{log}_a \space x^7y^8 - \text{log}_a \space \sqrt[4]{z}] \\[1em] \Rightarrow \dfrac{1}{3}[\text{log}_a \space x^7 + \text{log}_a \space y^8 - \text{log}_a \space z^{\dfrac{1}{4}}] \\[1em] \Rightarrow \dfrac{1}{3}[7\text{log}_a \space x + 8\text{log}_a \space y - \dfrac{1}{4}\text{log}_a \space z] \\[1em] \Rightarrow \dfrac{7}{3}\text{log}_a \space x + \dfrac{8}{3}\text{log}_a \space y - \dfrac{1}{12}\text{log}_a \space z.$

Hence, $\text{log}_a \space {\sqrt[3]{x^7y^8 ÷ \sqrt[4]{z}}} = \dfrac{7}{3}\text{log}_a \space x + \dfrac{8}{3}\text{log}_a \space y - \dfrac{1}{12}\text{log}_a \space z.$

Find the value of $\text{log}_{\sqrt{3}} \space 3\sqrt{3} - \text{log}_5 \space (0.04).$

Answer

Given,

$\Rightarrow \text{log}_{\sqrt{3}} \space 3\sqrt{3} - \text{log}_5 \space (0.04) \\[1em] \Rightarrow \text{log}_{\sqrt{3}} \space (\sqrt{3})^3 - \text{log}_5 \space \Big(\dfrac{4}{100}\Big) \\[1em] \Rightarrow 3\text{log}_{\sqrt{3}}\sqrt{3} - \text{log}_5\Big(\dfrac{1}{25}\Big) \\[1em] \Rightarrow 3(1) - \text{log}_5(5)^{-2} \\[1em] \Rightarrow 3 - (-2)\text{log}_55 \\[1em] \Rightarrow 3 + 2(1) \\[1em] \Rightarrow 5.$

Hence, $\text{log}_{\sqrt{3}} \space 3\sqrt{3} - \text{log}_5 \space (0.04)$ = 5.

Prove the following:

$(\text{log} \space x)^2 - (\text{log} \space y)^2 = \text{log} \space \dfrac{x}{y}.\text{log} \space xy$

Answer

Given,

$(\text{log} \space x)^2 - (\text{log} \space y)^2 = \text{log} \space \dfrac{x}{y}.\text{log} \space xy$

Simplifying L.H.S. of above equation we get,

$\Rightarrow (\text{log} \space x)^2 - (\text{log} \space y)^2 \\[1em] \Rightarrow (\text{log} \space x - \text{log} \space y)(\text{log} \space x + \text{log} \space y) \\[1em] \Rightarrow \text{log} \space\dfrac{x}{y}.\text{log} \space xy$

Since, L.H.S. = R.H.S.,

Hence, proved that $(\text{log} \space x)^2 - (\text{log} \space y)^2 = \text{log} \space \dfrac{x}{y}.\text{log} \space xy$

Prove the following:

$2\text{log} \space \dfrac{11}{13} + \text{log} \space \dfrac{130}{77} - \text{log} \space \dfrac{55}{91} = \text{log} \space 2.$

Answer

Given,

$2\text{log} \space \dfrac{11}{13} + \text{log} \space \dfrac{130}{77} - \text{log} \space \dfrac{55}{91} = \text{log} \space 2.$

Simplifying L.H.S. of the above equation we get,

$\Rightarrow 2\text{log} \space \dfrac{11}{13} + \text{log} \space \dfrac{130}{77} - \text{log} \space \dfrac{55}{91} \\[1em] \Rightarrow 2(\text{log} \space 11 - \text{log} \space 13) + (\text{log} \space 130 - \text{log} \space 77) - (\text{log} \space 55 - \text{log} \space 91) \\[1em] \Rightarrow 2(\text{log} \space 11 - \text{log} \space 13) + (\text{log} \space 13.10 - \text{log} \space 11.7) - (\text{log} \space 11.5 - \text{log} \space 13.7) \\[1em] \Rightarrow 2(\text{log} \space 11 - \text{log} \space 13) + (\text{log} \space 13 + \text{log} \space 10 - (\text{log} \space 11 + \text{log} \space 7) - (\text{log} \space 11 + \text{log} \space 5 - (\text{log} \space 13 + \text{log} \space 7))) \\[1em] \Rightarrow 2\text{log} \space 11 - 2\text{log} \space 13 + \text{log} \space 13 + \text{log} \space 10 - \text{log} \space 11 - \cancel{\text{log} \space 7} - \text{log} \space 11 - \text{log} \space 5 + \text{log} \space 13 + \cancel{\text{log} \space 7} \\[1em] \Rightarrow \cancel{2\text{log} \space 11} - \cancel{2\text{log} \space 11} + \cancel{2\text{log} \space 13} - \cancel{2\text{log} \space 13} + \text{log} \space 10 - \text{log} \space 5 \\[1em] \Rightarrow \text{log} \space 10 - \text{log} \space 5 \\[1em] \Rightarrow \text{log} \space 2.5 - \text{log} \space 5 \\[1em] \Rightarrow \text{log} \space 2 + \cancel{\text{log} \space 5} - \cancel{\text{log} \space 5} \\[1em] \Rightarrow \text{log} \space 2.$

Since, L.H.S. = R.H.S.,

Hence, proved that $2\text{log} \space \dfrac{11}{13} + \text{log} \space \dfrac{130}{77} - \text{log} \space \dfrac{55}{91} = \text{log} \space 2.$

If log (m + n) = log m + log n, show that n = $\dfrac{m}{m - 1}.$

Answer

Given,

log (m + n) = log m + log n

⇒ log (m + n) = log mn

⇒ m + n = mn

⇒ m = mn - n

⇒ m = n(m - 1)

⇒ n = $\dfrac{m}{m - 1}$.

Hence, proved that n = $\dfrac{m}{m - 1}$.

If $\text{log} \space \dfrac{x + y}{2} = \dfrac{1}{2}(\text{log} \space x + \text{log} \space y)$, prove that x = y.

Answer

Given,

$\Rightarrow \text{log} \space \dfrac{x + y}{2} = \dfrac{1}{2}(\text{log} \space x + \text{log} \space y) \\[1em] \Rightarrow \text{log} \space \dfrac{x + y}{2} = \dfrac{1}{2}(\text{log} \space xy) \\[1em] \Rightarrow \text{log} \space \dfrac{x + y}{2} = \text{log} \space (xy)^{\dfrac{1}{2}} \\[1em] \Rightarrow \dfrac{x + y}{2} = (xy)^{\dfrac{1}{2}} \\[1em] \Rightarrow x + y = 2(xy)^{\dfrac{1}{2}} \\[1em]$

Squaring both sides we get,

$\Rightarrow (x + y)^2 = 4(xy) \\[1em] \Rightarrow (x^2 + y^2 + 2xy) = 4xy \\[1em] \Rightarrow x^2 + y^2 + 2xy - 4xy = 0 \\[1em] \Rightarrow x^2 + y^2 - 2xy = 0 \\[1em] \Rightarrow (x - y)^2 = 0 \\[1em] \Rightarrow x - y = 0 \\[1em] \Rightarrow x = y.$

Hence, proved that x = y.

If a, b are positive real numbers, a > b and a² + b² = 27ab, prove that

$\text{log} \space\Big(\dfrac{a - b}{5}\Big) = \dfrac{1}{2}(\text{log} \space a + \text{log} \space b)$

Answer

Given,

a² + b² = 27ab

⇒ a² + b² = 2ab + 25ab

⇒ a² + b² - 2ab = 25ab

$\Rightarrow ab = \dfrac{a^2 + b^2 - 2ab}{25} = \Big(\dfrac{a - b}{5}\Big)^2$

Taking log on both sides:

$\Rightarrow \text{log} \space ab = \text{log} \space \Big(\dfrac{a - b}{5}\Big)^2 \\[1em] \Rightarrow \text{log} \space a + \text{log} \space b = 2 \text{log} \space \Big(\dfrac{a - b}{5}\Big) \\[1em] \Rightarrow \text{log} \space \Big(\dfrac{a - b}{5}\Big) = \dfrac{1}{2}(\text{log} \space a + \text{log} \space b).$

Hence, proved that $\text{log} \space \Big(\dfrac{a - b}{5}\Big) = \dfrac{1}{2}(\text{log} \space a + \text{log} \space b).$

Solve the following equation for x:

log_x $\dfrac{1}{49}$ = -2

Answer

Given,

$\Rightarrow \text{log}_x \space \dfrac{1}{49} = -2 \\[1em] \Rightarrow \dfrac{1}{7^2} = x^{-2} \\[1em] \Rightarrow 7^{-2} = x^{-2} \\[1em] \Rightarrow x = 7.$

Hence, x = 7.

Solve the following equation for x:

$\text{log}_x \space \dfrac{1}{4\sqrt{2}} = -5$

Answer

Given,

$\Rightarrow \text{log}_x \space \dfrac{1}{4\sqrt{2}} = -5 \\[1em] \Rightarrow \text{log}_x \space \dfrac{1}{2^2.2^{\dfrac{1}{2}}} = -5 \\[1em] \Rightarrow -\dfrac{1}{5}\text{log}_x \space \dfrac{1}{2^{2 + \frac{1}{2}}} = 1 \\[1em] \Rightarrow -\dfrac{1}{5}\text{log}_x \space \dfrac{1}{2^{\dfrac{5}{2}}} = 1 \\[1em] \Rightarrow -\dfrac{1}{5}\text{log}_x \space 2^{-\dfrac{5}{2}} = 1 \\[1em] \Rightarrow -\dfrac{1}{5} \times -\dfrac{5}{2} \text{log}_x \space 2 = 1 \\[1em] \Rightarrow \dfrac{1}{2}\text{log}_x \space 2 = 1 \\[1em] \Rightarrow \text{log}_x \space 2 = 2 \\[1em] \Rightarrow x^2 = 2 \\[1em] \Rightarrow x = \sqrt{2}.$

Hence, x = $\sqrt{2}$.

Solve the following equation for x:

$\text{log}_x \space \dfrac{1}{243} = 10$

Answer

Given,

$\Rightarrow \text{log}_x \space \dfrac{1}{243} = 10 \\[1em] \Rightarrow \text{log}_x \space 1 - \text{log}_x \space 243 = 10 \\[1em] \Rightarrow 0 - \text{log}_x \space (3)^5 = 10 \\[1em] \Rightarrow -5\text{log}_x \space 3 = 10 \\[1em] \Rightarrow -\text{log}_x \space 3 = \dfrac{10}{5} \\[1em] \Rightarrow \text{log}_x \space 3 = -2 \\[1em] \Rightarrow x^{-2} = 3 \\[1em] \Rightarrow \dfrac{1}{x^2} = 3 \\[1em] \Rightarrow \dfrac{1}{x} = \sqrt{3} \\[1em] \Rightarrow x = \dfrac{1}{\sqrt{3}}.$

Hence, x = $\dfrac{1}{\sqrt{3}}$.

Solve the following equation for x:

log₄ 32 = x - 4

Answer

Given,

⇒ log₄ 32 = x - 4

⇒ 4^{x - 4} = 32

⇒ (2²)^{x - 4} = 2⁵

⇒ 2^{2x - 8} = 2⁵

⇒ 2x - 8 = 5

⇒ 2x = 13

⇒ x = $\dfrac{13}{2} = 6\dfrac{1}{2}$.

Hence, x = $6\dfrac{1}{2}$.

Solve the following equation for x:

log₇ (2x² - 1) = 2

Answer

Given,

⇒ log₇ (2x² - 1) = 2

⇒ 2x² - 1 = 7²

⇒ 2x² - 1 = 49

⇒ 2x² = 50

⇒ x² = 25

⇒ x = 5, -5.

Hence, x = 5, -5.

Solve the following equation for x:

log (x² - 21) = 2

Answer

Given,

⇒ log (x² - 21) = 2

⇒ x² - 21 = 10²

⇒ x² = 100 + 21

⇒ x² = 121

⇒ x = 11, -11.

Hence, x = 11, -11.

Solve the following equation for x:

log₆ (x - 2)(x + 3) = 1

Answer

Given,

⇒ log₆ (x - 2)(x + 3) = 1

⇒ (x - 2)(x + 3) = 6¹

⇒ (x² + 3x - 2x - 6) = 6

⇒ x² + x - 6 = 6

⇒ x² + x - 12 = 0

⇒ x² + 4x - 3x - 12 = 0

⇒ x(x + 4) - 3(x + 4) = 0

⇒ (x - 3)(x + 4) = 0

⇒ x - 3 = 0 or x + 4 = 0

⇒ x = 3 or x = -4.

Hence, x = 3, -4.

Solve the following equation for x:

log₆ (x - 2) + log₆ (x + 3) = 1

Answer

Given,

⇒ log₆ (x - 2) + log₆ (x + 3) = 1

⇒ log₆ (x - 2)(x + 3) = 1

⇒ (x - 2)(x + 3) = 6¹

⇒ (x² + 3x - 2x - 6) = 6

⇒ x² + x - 6 = 6

⇒ x² + x - 12 = 0

⇒ x² + 4x - 3x - 12 = 0

⇒ x(x + 4) - 3(x + 4) = 0

⇒ (x - 3)(x + 4) = 0

⇒ x - 3 = 0 or x + 4 = 0

⇒ x = 3 or x = -4.

In this case x ≠ -4 as (x + 3) and (x - 2) will be negative and log of only positive numbers are defined.

Hence, x = 3.

Solve the following equation for x:

log (x + 1) + log (x - 1) = log 11 + 2log 3

Answer

Given,

⇒ log (x + 1) + log (x - 1) = log 11 + 2log 3

⇒ log (x + 1)(x - 1) = log 11 + log 3²

⇒ log (x² - 1) = log (11 x 9)

⇒ log (x² - 1) = log 99

⇒ (x² - 1) = 99

⇒ x² = 99 + 1

⇒ x² = 100

⇒ x = 10, -10

In this case x ≠ -10 as (x + 1) and (x - 1) will be negative and log of only positive numbers are defined.

Hence, x = 10.

Solve for x and y:

$\dfrac{\text{log} \space x}{3} = \dfrac{\text{log} \space y}{2}$ and $\text{log} \space (xy) = 5$.

Answer

Given,

$\dfrac{\text{log} \space x}{3} = \dfrac{\text{log} \space y}{2}$

⇒ 2 log x = 3 log y

⇒ 2 log x - 3 log y = 0 .......(i)

Given,

log (xy) = 5

⇒ log x + log y = 5 ......(ii)

Multiplying (ii) by 2 we get,

⇒ 2 log x + 2 log y = 10 ......(iii)

Subtracting (i) from (iii) we get,

⇒ 2 log x + 2 log y - (2 log x - 3 log y) = 10 - 0

⇒ 2 log x - 2 log x + 2 log y + 3 log y = 10

⇒ 5 log y = 10

⇒ log y = 2

⇒ y = 10² = 100.

Substituting value of y in (ii) we get,

⇒ log x + log 100 = 5

⇒ log x + 2 = 5

⇒ log x = 3

⇒ x = 10³ = 1000.

Hence, x = 1000 and y = 100.

If a = 1 + log_x yz, b = 1 + log_y zx and c = 1 + log_z xy, then show that

ab + bc + ca = abc.

Answer

a = 1 + log_x yz = log_x x + log_x yz = log_x xyz.

$\therefore \dfrac{1}{a} = \text{log}_{xyz} \space x$

b = 1 + log_y zx = log_y y + log_y zx = log_y xyz.

$\therefore \dfrac{1}{b} = \text{log}_{xyz} \space y$

c = 1 + log_z xy = log_z z + log_z xy = log_z xyz.

$\therefore \dfrac{1}{c} = \text{log}_{xyz} \space z$

$\Rightarrow \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \text{log}_{xyz} \space x + \text{log}_{xyz} \space y + \text{log}_{xyz} \space z \\[1em] \Rightarrow \dfrac{bc + ac + ab}{abc} = \dfrac{\text{log x}}{\text{log xyz}} + \dfrac{\text{log y}}{\text{log xyz}} + \dfrac{\text{log z}}{\text{log xyz}} \\[1em] \Rightarrow \dfrac{bc + ac + ab}{abc} = \dfrac{\text{log x + log y + log z}}{\text{log xyz}} \\[1em] \Rightarrow \dfrac{bc + ac + ab}{abc} = \dfrac{\text{log xyz}}{\text{log xyz}} \\[1em] \Rightarrow \dfrac{bc + ac + ab}{abc} = 1 \\[1em] \Rightarrow ab + bc + ca = abc.$

Hence, proved that ab + bc + ca = abc.

If $\dfrac{1}{\text{ log }_a x} + \dfrac{1}{\text{ log }_b x} = \dfrac{2}{\text{ log }_c x}$, prove that c² = ab.

Answer

Given,

$\Rightarrow \dfrac{1}{\text{ log }_a x} + \dfrac{1}{\text{ log }_b x} = \dfrac{2}{\text{ log }_c x}\\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{log }x}{\text{log }a}} + \dfrac{1}{\dfrac{\text{log } x}{\text{log } b}} = \dfrac{2}{\dfrac{\text{log }x}{\text{log }c}} \\[1em] \Rightarrow \dfrac{\text{ log } a}{\text{ log } x} + \dfrac{\text{ log } b}{\text{ log } x} = \dfrac{2\text{ log } c}{\text{ log } x}\\[1em] \Rightarrow \dfrac{\text{ log } a + \text{ log } b}{\text{ log } x} = \dfrac{2\text{ log } c}{\text{ log } x}\\[1em] \Rightarrow \dfrac{\text{ log } (a \times b)}{\text{ log } x} = \dfrac{2\text{ log } c}{\text{ log } x}\\[1em] \Rightarrow \dfrac{\text{ log } ab}{\text{ log } x} = \dfrac{2\text{ log } c}{\text{ log } x}\\[1em] \Rightarrow \text{ log } ab = 2\text{ log } c\\[1em] \Rightarrow \text{ log } ab = \text{ log } c^2\\[1em] \Rightarrow ab = c^2.$

Hence, proved that c² = ab.