Question

Find the value of X, when 2^(x+4) × 3^(x+1) = 288​

Answer 1

$\large\text{\underline{Question:-}}$

Solve the exponential equation

$\hookrightarrow \boxed{2^{x+4}\cdot 3^{x+1}=288}$

$\large\text{\underline{Solution:-}}$

Note that,

$\hookrightarrow 2^{x+4}=2^{4}\cdot 2^{x}\text{ and }3^{x+1}=3\cdot 3^{x}$

Now, from the above,

$\hookrightarrow 2^{4}\cdot 3\cdot2^{x}\cdot 3^{x}=288$

$\hookrightarrow 2^{x}\cdot 3^{x}=\dfrac{2^{5}\cdot 3^{2}}{2^{4}\cdot 3}$

We know that,

$\hookrightarrow \boxed{\dfrac{a^{x}}{a^{y}}=a^{x-y} }$

$\hookrightarrow 2^{x}\cdot 3^{x}=2^{5-4}\cdot3^{2-1}$

$\hookrightarrow 2^{x}\cdot 3^{x}=6$

We know that,

$\hookrightarrow \boxed{a^{x}b^{x}=(ab)^{x}}$

So,

$\hookrightarrow 6^{x}=6^{1}$

Now we compare the exponent as the base is the same.

$\hookrightarrow x=1$

$\large\text{\underline{Verification:-}}$

$\hookrightarrow \text{(L.H.S)}=2^{4+1}\cdot 3^{1+1}$

$\hookrightarrow \text{(L.H.S)}=2^{5}\cdot 3^{2}$

$\hookrightarrow \text{(R.H.S)}=288$

$\hookrightarrow \text{(R.H.S)}=2^{5}\cdot 3^{2}$

Hence,

$\hookrightarrow \text{(L.H.S)}=\text{(R.H.S)}$

$\large\text{\underline{Answer:-}}$

This equation has $x=1$ as a solution.

Answer 2

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