Question

Find m so that (2/9)^3 × (2/9)^6 = (2/9)^2m-1

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Answer 1

☞ Solution

= 3+6 = 2m-1

9 = 2m-1

9+1 = 2m

10 = 2m

m = 10/2

m = 5

Hope it helps uh Ãmî!

Answer 2

$\underline{ \underline{ \Large \pmb{\sf { {Given \: Equation:}} }} }$

$\bigstar \: \boxed{\sf {{\Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{3} \times {\Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{6} = {\Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{2m -1} }}$

$\underline{ \underline{ \Large \pmb{\sf { {To \: calculate:}} }} }$

• Value of m.

$\underline{ \underline{ \Large \pmb{\sf { {Solution:}} }} }$

Remember:

» Whenever we're solving any questions related to exponents, it should be kept in mind that if the bases are equal in L.H.S and R.H.S , then their powers must be equal in L.H.S and R.H.S. So, to find the value of m, we should note the powers of the given numbers as here, bases are equal .

$\longrightarrow \sf {{ \Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{3+6} = {\Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{2m -1} }$

$\longrightarrow \sf {{ \Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{9} = {\Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{2m -1} }$

Since, the bases are equal , their powers must be equal. So,

$\longrightarrow \sf { 9 = 2m - 1 }$

$\longrightarrow \sf { 9 + 1 = 2m }$

$\longrightarrow \sf { 10 = 2m }$

$\longrightarrow \sf { \cancel{ \dfrac{10}{2}} = m }$

$\longrightarrow \boxed { \pmb { \rm \red { 5 = m }}}$

Therefore, value of m is 5.

$\underline{ \underline{ \Large \pmb{\sf { {Verification:}} }} }$

$\bigstar \: \boxed{\sf {{\Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{3} \times {\Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{6} = {\Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{2m -1} }}$

Let us substitute the value of m in R.H.S to check whether L.H.S and R.H.S are equal or not.

L.H.S :

$\longrightarrow \sf {{ \Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{3} \times { \Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{3} }$

$\longrightarrow \sf {{ \Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{3+6} }$

$\longrightarrow \boxed { \pmb { \rm \green{{ \Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{9} }}}$

R.H.S :

$\longrightarrow \sf {{ \Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{2m - 1} }$

$\longrightarrow \sf {{ \Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{2(5) - 1} }$

$\longrightarrow \sf {{ \Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{10- 1} }$

$\longrightarrow \boxed { \pmb { \rm \green{ {\Bigg \lgroup \dfrac{2}{9} \Bigg \rgroup }^{9} }}}$

Therefore, L.H.S = R.H.S. Hence , verified!!