Solve each equation. Use one of the 4 methods you have practiced the last few days:
1. Write exponents using the same base
2. Take the log of both sides
3. Use properties of logs then write as an exponent
4. Use properties of logs then drop the log on both sides
1. 4^-x =32

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msm555 msm555 · Answer 1 · 2024-03-12T15:02:29+01:00

Answer:

[tex]x = -\dfrac{5}{2}[/tex]

Step-by-step explanation:

To solve the equation [tex]4^{-x} = 32[/tex], let's use method 1 (Write exponents using the same base).

[tex] 4^{-x} = 32 [/tex]

Since [tex]32 = 2^5[/tex], we can rewrite [tex]32[/tex] as [tex]2^5[/tex]:

[tex] 4^{-x} = 2^5 [/tex]

Now, we notice that [tex]4 = 2^2[/tex], so we can rewrite [tex]4[/tex] as [tex]2^2[/tex]:

[tex] (2^2)^{-x} = 2^5 [/tex]

Now, apply the power of a power rule which states [tex](a^m)^n = a^{mn}[/tex]:

[tex] 2^{-2x} = 2^5 [/tex]

Now, since the bases are the same, we can equate the exponents:

[tex] -2x = 5 [/tex]

Now, solve for [tex]x[/tex]:

[tex] x = \dfrac{5}{-2} [/tex]

[tex] x = -\dfrac{5}{2} [/tex]

So, the solution to the equation is:

[tex]x = -\dfrac{5}{2}[/tex]