Question

find the value of x in (1/3)^3×(1/3)^(-6)=(1/3)^(2x-1)

Answer 1

Answer:

we given

(1/3)^3 * (1/3)^(-6) = (1/3)^(2x-1)

(1/3)^(-3) = (1/3)^(2x-1)

comparing both sides, we get

-3 = 2x -1

-3+1 = 2x

-2 = 2x

x = -1

Answer 2

$\begin{gathered} \begin{gathered}\begin{gathered}\\\;\underbrace{\underline{\sf{Understanding\; the \; Question:-}}}\end{gathered}\end{gathered} \end{gathered}$

Here the Concept of exponents and power has been used. Here we see that the bases are same so by equating exponents of both sides we can get the answer .

Let's do it !!

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★ Question :-

find the value of x in

$\tt \bigg(\dfrac{1}{3} \bigg) ^{3} \times \bigg( \dfrac{1}{3} \bigg) ^{ - 6} = \bigg(\dfrac{1}{3} \bigg) ^{2x - 1}$

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★ Solution :-

$\begin{gathered}\\\;\bf{\rightarrow\;\;\green{\bigg(\dfrac{1}{3} \bigg) ^{3} \times \bigg( \dfrac{1}{3} \bigg) ^{ - 6} = \bigg(\dfrac{1}{3} \bigg) ^{2x - 1} }}\end{gathered}$

So we can see that all the bases are same so , by equating exponents of both side we get ,

$\begin{gathered}\\\;\bf{\rightarrow\;\;\purple{ {3} - { 6} = {2x - 1} }}\end{gathered}$

$\begin{gathered}\\\;\bf{\rightarrow\;\;{ { - 3} + 1 = 2x }}\end{gathered}$

$\begin{gathered}\\\;\bf{\rightarrow\;\;{ - 2= 2x }}\end{gathered}$

$\begin{gathered}\\\;\bf{\rightarrow\;\;{ x= \dfrac{ - 2}{2} }}\end{gathered}$

$\begin{gathered}\\\;\bf{\rightarrow\;\;\blue{ - 1 }}\end{gathered}$

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Let's check that our answer is correct or incorrect !

Given equation

$\begin{gathered}\\\;\bf{\rightarrow\;\;\red{\bigg(\dfrac{1}{3} \bigg) ^{3} \times \bigg( \dfrac{1}{3} \bigg) ^{ - 6} = \bigg(\dfrac{1}{3} \bigg) ^{2x - 1} }}\end{gathered}$

after putting -1 in the place of x we get

$\begin{gathered}\\\;\bf{\rightarrow\;\;\orange{\bigg(\dfrac{1}{3} \bigg) ^{3} \times \bigg( \dfrac{1}{3} \bigg) ^{ - 6} = \bigg(\dfrac{1}{3} \bigg) ^{2 \times (- 1)- 1} }}\end{gathered}$

$\begin{gathered}\\\;\bf{\rightarrow\;\;{\bigg(\dfrac{1}{3} \bigg) ^{3} \times \bigg( \dfrac{1}{3} \bigg) ^{ - 6} = \bigg(\dfrac{1}{3} \bigg) ^{ - 2- 1} }}\end{gathered}$

$\begin{gathered}\\\;\bf{\rightarrow\;\;{\bigg(\dfrac{1}{3} \bigg) ^{3} \times \bigg( \dfrac{1}{3} \bigg) ^{ - 6} = \bigg(\dfrac{1}{3} \bigg) ^{ - 3 } }}\end{gathered}$

Now let's make the power positive ,

$\begin{gathered}\\\;\bf{\rightarrow\;\;{\bigg(\dfrac{1}{3} \bigg) ^{3} \times \bigg( \dfrac{3}{1} \bigg) ^{6} = \bigg(\dfrac{3}{1} \bigg) ^{ 3 } }}\end{gathered}$

Now we want to check that our answer is correct or not therefore first we will calculate the value of LHS and then the value of RHS

Therefore ,

LHS =

$\begin{gathered}\\\;\bf{\rightarrow\;\;{ \pink{\bigg (\dfrac{1}{3} \bigg) ^{3} \times \bigg( \dfrac{3}{1} \bigg) ^{6} } }}\end{gathered}$

$\begin{gathered}\\\;\bf{\rightarrow\;\;{\bigg (\dfrac{1}{27} \bigg) \times \bigg( \dfrac{729}{1} \bigg) } }\end{gathered}$

$\begin{gathered}\\\;\bf{\rightarrow\;\;{\bigg (\dfrac{1}{ \cancel{27}} \bigg) \times \bigg( \dfrac{ \cancel{729}}{1} \bigg) }= \green{} \green{ \dfrac{27}{1} } }\end{gathered}$

RHS =

$\begin{gathered}\\\;\bf{\rightarrow\;\;{ \pink{\bigg (\dfrac{3}{1} \bigg) ^{3}} }}\end{gathered}$

$\begin{gathered}\\\;\bf{\rightarrow\;\;{ {\bigg (\dfrac{3 \times 3 \times 3}{1 \times 1 \times 1} \bigg) = \green{ \dfrac{27}{1} }} }}\end{gathered}$

hence ,

$\begin{gathered}\\\;\bf{\rightarrow\;\;{ { \green{ \dfrac{27}{1} = \dfrac{27}{1} }} }}\end{gathered}$

LHS = RHS , hence our answer is correct .

$\: \: \boxed{\boxed{\bf{\mapsto \: \: \: hence , The \; value \: of \: x \: = \blue{\underline{-1}}}}}$