Question

rmula1" title="4 \times 8 ^{2x + 1?} = 32 ^{x + 2} " alt="4 \times 8 ^{2x + 1?} = 32 ^{x + 2} " align="absmiddle" class="latex-formula">
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Answer 1

Answer:

x = 5

Step-by-step explanation:

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FACTS TO KNOW BEFORE SOLVING :-

• $a^x \times a^y = a^{x+y}$

• In an equation , if the bases are same in both L.H.S. & R.H.S. then , the power of the bases on both the sides of equation should be equal. For e.g. : $a^x = a^y$  ⇒  $x = y$  [∵ Bases are equal on both the sides]

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$4 \times 8^{2x+1} = 32^{x+2}$

Lets express it in terms of 2.

$=> 2^2 \times 2^{3 (2x+1)} = 2^{5(x+2)}$

$=> 2^2 \times 2^{6x+3} = 2^{5x+10}$

$=> 2^{2 + 6x+3} = 2^{5x+10}$

$=> 2^{6x+5} = 2^{5x+10}$

Here the bases on both the sides are equal. Hence ,

$=> 6x + 5 = 5x + 10$

$=> 6x - 5x = 10 - 5$

$=> x = 5$