Question

1. Find the valu of x
$\sf {8}^{2x + 2} = {16}^{3x - 1}$
2. Evaluate :-
$\sf\frac{36^{ \frac{5}{2}} - 36^{ \frac{7}{2} } }{36^{ \frac{3}{2} } }$
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Answer 1

Given :

1. Find the value of x

$\sf {8}^{2x + 2} = {16}^{3x - 1}$

2. Evaluate :-

$\sf\frac{36^{ \frac{5}{2}} - 36^{ \frac{7}{2} } }{36^{ \frac{3}{2} } }$

Solution :

$\sf1) \: {8}^{2x + 2} = {16}^{3x - 1}$

8 can be written as 2³ and 16 can be written as 2⁴

So, the equation is :

$\sf {2}^{3(2x + 2)} = {2}^{4(3x - 1)}$

$\sf {2}^{6x + 6} = {2}^{12x - 4}$

As the bases are same they will be eliminated

Now, we get

$\sf 6x + 6 = 12x - 4$

By transposing the numbers we get

$\sf6x - 12x = - 4 - 6$

$\sf - 6x = - 10$

(-) will be eliminated

$\sf6x = 10$

$\sf x = \frac{10}{6}$

$\sf x = \frac{5}{3}$

Therefore, the value of x is 5/3

$\sf2) \:\frac{36^{ \frac{5}{2}} - 36^{ \frac{7}{2} } }{36^{ \frac{3}{2} } }$

Here we can write 36 in the exponential form which 6²

$\frac{ {6}^{ \cancel2 (\frac{5}{ \cancel2} )} - {6}^{ \cancel2( \frac{7}{ \cancel2} )} }{ {6}^{ \cancel2( \frac{3}{ \cancel2} )} }$

So, we got

$\frac{ {6}^{5} - {6}^{7} }{ {6}^{3} }$

By taking 6³ we get

${6}^{5} - {6}^{7} \div {6}^{3}$

Bases are same

${6}^{5} - {6}^{(7-3)}$

${6}^{5} - {6}^{4}$

7776 - 1296

6480

Henceforth, the answer is 6480