Question

please answer me..... ​

Answer 1

Required Answer:-

Given:

$\rm \mapsto {a}^{l} = {b}^{m} = {e}^{n} = {d}^{p}$

To prove:

$\rm \mapsto log_{a}(bed) = l \bigg( \dfrac{1}{m} + \dfrac{1}{n} + \dfrac{1}{p}\bigg)$

Proof:

Given that,

$\rm \mapsto {a}^{l} = {b}^{m} = {e}^{n} = {d}^{p}$

Now,

$\rm \implies {b}^{m} = {a}^{l}$

$\rm \implies b= {a}^{^{l}/ \: _{m}}$

Also,

$\rm \implies {e}^{n} = {a}^{l}$

$\rm \implies e= {a}^{^{l}/ \: _{n}}$

Again,

$\rm \implies {d}^{p} = {a}^{l}$

$\rm \implies d= {a}^{^{l}/ \: _{p}}$

Now, Taking LHS,

$\rm log_{a}(bed)$

$\rm = log_{a} \bigg( {a}^{ \dfrac{l}{m} } \times {a}^{ \dfrac{l}{n} } \times {a}^{ \dfrac{l}{p} } \bigg)$

$\rm = log_{a} \bigg( {a}^{ \dfrac{l}{m} + \dfrac{l}{n} + \dfrac{l}{p} } \bigg)$

$\rm = log_{a} ( {a})^{ \bigg({ \dfrac{l}{m} + \dfrac{l}{n} + \dfrac{l}{p} \bigg )}}$

$\rm = { \bigg({ \dfrac{l}{m} + \dfrac{l}{n} + \dfrac{l}{p} \bigg )}} \times \cancel{log_{a}(a)}$

$\rm = { \bigg({ \dfrac{l}{m} + \dfrac{l}{n} + \dfrac{l}{p} \bigg )}}$

$\rm = l{ \bigg({ \dfrac{1}{m} + \dfrac{1}{n} + \dfrac{1}{p} \bigg )}}$

= RHS (Hence Proved)

Note:

$\rm \mapsto log_{x}(x) = 1 \: \: (x \neq 1)$

Answer 2

Answer:

In physics, a force is any interaction that, when unopposed, will change the motion of an object. A force can cause an object with mass to change its velocity, i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity