Answer:
[tex] x = 2.25 [/tex]
Step-by-step explanation:
Let's solve the equation [tex]4^x = 16\sqrt{2}[/tex] using both methods:
Using Powers of the Same Number:
We need to express [tex]16\sqrt{2}[/tex] as a power of [tex]2[/tex].
Notice that:
[tex] 4^x = 2^{2x}[/tex]
[tex]16\sqrt{2} = 2^4 \times (\sqrt{2}) = (2^4) \times (2^{\dfrac{1}{2}}) = 2^{4 + \dfrac{1}{2}} = 2^{\dfrac{9}{2}}[/tex].
So, [tex]2^{2x} = 4^{\dfrac{9}{2}}[/tex], then
[tex]2x = \dfrac{9}{2} [/tex]
[tex]x = \dfrac{9}{2\cdot 2 }[/tex]
[tex] x =\dfrac{9}{4}[/tex]
[tex] x = 2.25 [/tex]
Using Logarithms:
We rewrite the equation [tex]4^x = 16\sqrt{2}[/tex] in logarithmic form:
[tex] x = \log_4(16\sqrt{2}) [/tex]
Using properties of logarithms:
[tex] x = \log_4(16) + \log_4(\sqrt{2}) [/tex]
[tex] x = \log_4(16) + \dfrac{1}{2} \log_4(2) [/tex]
Now, [tex]16[/tex] is [tex]4^2[/tex], so [tex]\log_4(16) = 2[/tex]. And [tex]\sqrt{2}[/tex] is [tex]2^{\dfrac{1}{2}}[/tex], so [tex]\log_4(\sqrt{2}) = \dfrac{1}{2}[/tex].
[tex] x = 2 + \dfrac{1}{2} \times \dfrac{1}{2} [/tex]
[tex] x = 2 + \dfrac{1}{4} [/tex]
[tex] x =\dfrac{9}{4}[/tex]
[tex] x = 2.25 [/tex]
So, the value of x is 2.25.
