Question

Find the value of x if (1/5)-^8×(1/5)+⁴=(1/5)-⁴^x​

Answer 1

Solution:

$\bigg(\dfrac{1}{5} \bigg) ^{ - 8} \times \bigg(\dfrac{1}{5} \bigg) ^{4} = \bigg(\dfrac{1}{5} \bigg) ^{ - 4x}$

Here, we use this identity

$\pink \star \: \purple{ \rm{a^m \times a^n = a^{(m + n)}}}$

$\green{ \star} \: \red{ \sf{If \: a^{p} = a^{q}, \: Then \: p = q }}$

On further Simplifying,

$\dashrightarrow \: \: \bigg(\dfrac{1}{5} \bigg)^{ - 8 + 4} = \bigg(\dfrac{1}{5} \bigg)^{ - 4x}$

Using Second identity,

$\dashrightarrow \: \: \rm{ - 8 + 4 = - 4x}$

$\hookrightarrow \: \: \rm{ - 4 = - 4x}$

$\implies \boxed{ \bf{x = - 1}}$

∴ The value of x = -1

Additional information:

$\begin{gathered}\boxed{\begin{minipage}{5 cm}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{minipage}}\end{gathered}$

Do see this answer from brainly website (https://brainly.in/question/41493429) to see the additional information part clearly.