Question

$\tt{\underline{\boxed{\red{Question:-}}}}$
$\sf{If\:{40}^{a}\:=\:2,\:{40}^{b}\:=\:4,\:then\:find\:the\:value\:of}$
$\sf{{10}^{\Bigg\lgroup\dfrac{1\:-\:a\:-\:b}{2(1\:-\:b)}\Bigg\rgroup}}$​

Answer 1

AnswEr :

Given that,

$\sf \: {40}^{a} = 2 - - - - - - - (1) \\ \sf {40}^{b} = 4 - - - - - (2)$

We know that, 4 = 2²

$\implies \sf \: {40}^{b} = {2}^{2}$

Using (1),

$\implies \sf \: {40}^{b} = { {40}^{} }^{2a}$

Since, the bases are equal.

$\star \: \boxed{ \boxed{ \sf \: 2a = b}}$

Consider equation (1),

$\implies \sf \: {40}^{a} = {2}^{}$

Taking log on both sides of the equation,

$\longrightarrow \sf \: a log(40) = log(2) \\ \\ \longrightarrow \sf \: a = \dfrac{ log(2) }{ log(40) } \\ \\ \longrightarrow \sf \: a \sim \: 0.20$

Simplifying the exponent,

$\implies \sf \: \dfrac{1 - a - b}{2(1 - b)} \\ \\ \implies \sf \: \dfrac{1 - a - 2a}{2(1 - 2a)} \\ \\ \implies \sf \: \dfrac{1 - 3a}{2 - 4a} \\ \\ \implies \sf \: \dfrac{1 - 0.6}{2 - 0.8} \\ \\ \implies \sf \: \dfrac{0.4}{1.2} \\ \\ \implies \sf \: \dfrac{1}{3}$

Thus,

$\implies \sf \: {10}^{ \frac{1}{3} } \\ \\ \implies \sf \: 2.15$

Answer 2

Answer:

$\tt{\underline{\boxed{\red{Question:-}}}}$

$\sf{If\:{40}^{a}\:=\:2,\:{40}^{b}\:=\:4,\:then\:find\:the\:value\:of}$

$\sf{{10}^{\Bigg\lgroup\dfrac{1\:-\:a\:-\:b}{2(1\:-\:b)}\Bigg\rgroup}}$

AnswEr :

Given that,

$\begin{gathered}\sf \: {40}^{a} = 2 - - - - - - - (1) \\ \sf {40}^{b} = 4 - - - - - (2)\end{gathered}$

40

a

=2−−−−−−−(1)

40

b

=4−−−−−(2)

We know that, 4 = 2²

$\implies \sf \: {40}^{b} = {2}^{2}⟹40$

b

=2

2

Using (1),

$\implies \sf \: {40}^{b} = { {40}^{} }^{2a}⟹40$

b

=40

2a

Since, the bases are equal.

$\star \: \boxed{ \boxed{ \sf \: 2a = b}}⋆$

2a=b

Consider equation (1),

\implies \sf \: {40}^{a} = {2}^{}⟹40

a

=2

Taking log on both sides of the equation,

$\begin{gathered}\longrightarrow \sf \: a log(40) = log(2) \\ \\ \longrightarrow \sf \: a = \dfrac{ log(2) }{ log(40) } \\ \\ \longrightarrow \sf \: a \sim \: 0.20\end{gathered}$

⟶alog(40)=log(2)

⟶a=

log(40)

log(2)

⟶a∼0.20

Simplifying the exponent,

$\begin{gathered}\implies \sf \: \dfrac{1 - a - b}{2(1 - b)} \\ \\ \implies \sf \: \dfrac{1 - a - 2a}{2(1 - 2a)} \\ \\ \implies \sf \: \dfrac{1 - 3a}{2 - 4a} \\ \\ \implies \sf \: \dfrac{1 - 0.6}{2 - 0.8} \\ \\ \implies \sf \: \dfrac{0.4}{1.2} \\ \\ \implies \sf \: \dfrac{1}{3}\end{gathered}$

⟹

2(1−b)

1−a−b

⟹

2(1−2a)

1−a−2a

⟹

2−4a

1−3a

⟹

2−0.8

1−0.6

⟹

1.2

0.4

⟹

3

1

Thus,

[tex] \begin{gathered}\implies \sf \: {10}^{ \frac{1}{3} } \\ \\ \implies \sf \: 2.15\end{gathered} [/tex}

⟹10

3

1

⟹2.15