 QUESTION

# Find the value of x if the given equation for x is written as ${4^{2x - 1}} - {16^{x - 1}} = 384$.

Hint: To find the value of x we had to write 16 as ${4^2}$ in the given equation. And take ${4^{2x}}$ common from the LHS of the equation. After that equating the powers of both sides of the equation we will get the required value of x.

As we know that 16 is also written as ${4^2}$.
And the given equation is,
${4^{2x - 1}} - {16^{x - 1}} = 384$ ----(1)

Now replacing 16 with ${4^2}$ in equation 1. We get,
${4^{2x - 1}} - {\left( 4 \right)^{2(x - 1)}} = 384$
${4^{2x - 1}} - {\left( 4 \right)^{2x - 2}} = 384$ ----(2)

Now as we know that if ${a^{cx + d}}$ is any number than it can also be written as ${a^{cx}} \times {a^d}$
And if ${a^{cx - d}}$ is a number than it can be written as $\dfrac{{{a^{cx}}}}{{{a^d}}}$.
So, using above stated identity equation 2 can also be written as, $\dfrac{{{4^{2x}}}}{4} - \dfrac{{{4^{2x}}}}{{{{\left( 4 \right)}^2}}} = 384$
Now taking ${4^{2x}}$ common from the LHS of the above equation and then taking LCM on the LHS of the above equation. We get,
${4^{2x}}\left( {\dfrac{1}{4} - \dfrac{1}{{{{\left( 4 \right)}^2}}}} \right) = 384 \\ {4^{2x}}\left( {\dfrac{1}{4} - \dfrac{1}{{16}}} \right) = 384 \\ {4^{2x}}\left( {\dfrac{{16 - 4}}{{4 \times 16}}} \right) = {4^{2x}}\left( {\dfrac{{12}}{{64}}} \right) = {4^{2x}}\left( {\dfrac{3}{{16}}} \right) = 384 \\$
Now multiplying both sides of the above equation by $\dfrac{{16}}{3}$. We get,
${4^{2x}} = 384 \times \dfrac{{16}}{3} = 128 \times 16 = 64 \times 2 \times 16$
${4^{2x}} = 64 \times 2 \times 16$ ----(3)

Now as we know that 64 can be written as ${4^3}$, 16 can be written as ${4^2}$ and 2 is the square root of 4. So, 2 can also be written as ${4^{\dfrac{1}{2}}}$.
So, above equation 3 can also be written as,
${4^{2x}} = {4^3} \times {4^{\dfrac{1}{2}}} \times {4^2}$ -----(4)

As we know that if the numbers are multiplied like ${a^c} \times {a^d}$, then their powers are added. So, ${a^c} \times {a^d} = {a^{c + d}}$
So, equation 4 becomes,
${4^{2x}} = {4^{\left( {3 + \dfrac{1}{2} + 2} \right)}} = {4^{\dfrac{{11}}{2}}}$ -----(5)

Now as we know that if the equation is ${a^c} = {a^d}$ then c must be equal to d.
So, from equation 5 we can also write,
2x = $\dfrac{{11}}{2}$
So, dividing both sides of the above equation by 2. We get,
x = $\dfrac{{11}}{4}$
Hence, the required value of x will be equal to $\dfrac{{11}}{4}$.

Note: Whenever we come with this type of problem then first, we have to change the equation such that the number which has all numbers raised to the power becomes equal. Like here we change 16 to 4. And then we take common terms out and solve the equation. Like here the common part is ${4^{2x}}$. And then we change the other side of the other side of the equation as ${4^a}$. Where a is any number like here a = $\dfrac{{11}}{2}$. And then we equate powers of both sides of the equation to get the required value of x.