Question

BE HAPPY ,NO SPAM CLEAR​

Answer 1

1) Question:-

$log_{ \sqrt{5} }125$

Solution:-

Let

$log_{ \sqrt{5} }125 = x$

So,

$( \sqrt{5} )^{x} = 125$

$125 = {5}^{3} = ( { \sqrt{5} )}^{6}$

$( \sqrt{ {5}}) ^{x} = ( { \sqrt{5} })^{6}$

Bases are equal. So, powers also equal

$x = 6$

$log_{ \sqrt{5} }125 = 6$

2) Question:-

$log_{(0.25)}32$

Solution:-

Let

$log_{ (\frac{1}{4}) }32 = x$

So,

${( \frac{1}{4}) }^{x} = 32$

$\frac{1}{4} = \frac{1}{ {2}^{2} } = {2}^{ - 2}$

So,

${2}^{ - 2x} = 32$

$32 = 2 \times 2 \times 2 \times 2 \times 2 = {2}^{5}$

${2}^{ - 2x} = {2}^{5}$

Bases are equal. So, powers also equal

$- 2x = 5$

$x = - \frac{ 5}{2}$

$log_{(0.25)}32 = - \frac{5}{2}$

3) Question:-

$log_{( \frac{1}{8} )}\sqrt{32}$

Solution:-

Let

$log_{ \frac{1}{8} } \sqrt{32} = x$

$( { \frac{1}{8}) }^{x} = \sqrt{32}$

$\frac{1}{8} = \frac{1}{ {2}^{3} } = {2}^{ - 3}$

$\sqrt{32} = {32}^{ - 2} = {2}^{(5 \times - 2)}$

${2}^{ - 3x} = {2}^{ - 10}$

Bases are equal. So, powers also equal

$- 3x = - 10$

$x = \frac{ - 10}{ - 3} = \frac{10}{3}$

$log_{ \frac{1}{8} } \sqrt{32} = \frac{10}{3}$

Hope this is helpful✓

Answer 2

hope helpful to you please make me bralinist votes