Answer:
x = -6.5
Step-by-step explanation:
To solve the equation [tex]\sf \left(\dfrac{1}{16}\right)^{x+5} = 8^2[/tex], let's use the logarithmic property that states:
[tex]\sf a^{\log_a(b)} = b[/tex]
Here, [tex]\sf a = \frac{1}{16}[/tex], [tex]\sf b = 8^2[/tex], and [tex]\sf x + 5[/tex] is the exponent.
So, we can rewrite the equation as:
[tex]\sf \left(\frac{1}{16}\right)^{x+5} = 8^2[/tex]
[tex]\sf x + 5 = \log_{\frac{1}{16}}(8^2)[/tex]
Now, calculate [tex]\sf \log_{\frac{1}{16}}(8^2)[/tex]:
[tex]\sf x + 5 = \log_{\frac{1}{16}}(64)[/tex]
We know that [tex]\sf \frac{1}{16} = (16)^{-1}[/tex], so:
[tex]\sf x + 5 = \log_{(16)^{-1}}(64)[/tex]
Using the property [tex]\sf \log_b(a^c) = c \cdot \log_b(a)[/tex]:
[tex]\sf x + 5 = -\log_{16}(64)[/tex]
[tex]\sf x + 5 = -\log_{16}(2^6)[/tex]
[tex]\sf x + 5 = -6 \cdot \log_{16}(2)[/tex]
Now, we know that [tex]\sf 16 = 2^4[/tex], so:
[tex]\sf x + 5 = -6 \cdot \log_{2^4}(2)[/tex]
[tex]\sf x + 5 = -6 \cdot \dfrac{1}{4}[/tex]
[tex]\sf x + 5 = -\dfrac{3}{2}[/tex]
Subtract 5 from both sides:
[tex]\sf x = -\dfrac{3}{2} - 5[/tex]
[tex]\sf x = -\dfrac{13}{2}[/tex]
[tex] \sf x = - 6.5 [/tex]
Ttherefore, x = -6.5
