We must solve the following equation:
[tex]8^{x+2}=4.[/tex]
1) First, we express the 8 and the 4 as a powers of 2:
[tex]\begin{gathered} 8=2^3, \\ 4=2^2\text{.} \end{gathered}[/tex]
Replacing these equations in the equation above, we have:
[tex](2^3)^{x+2}=2^2.[/tex]
2) Using the property that the exponents multiplies we have:
[tex]2^{3\cdot(x+2)}=2^2.[/tex]
3) Because the basis are equal, the exponents must be equal too, so:
[tex]3\cdot(x+2)=2.[/tex]
4) Finally, we solve the last equation for x:
[tex]\begin{gathered} 3\cdot(x+2)=2, \\ 3x+6=2, \\ 3x=2-6, \\ 3x=-4, \\ x=-\frac{4}{3}\text{.} \end{gathered}[/tex]
Answer: x = -4/3
Summary:
[tex]\begin{gathered} 8^{x+2}=4 \\ \Leftrightarrow(2^3)^{x+2}=2^2 \\ \Leftrightarrow2^{3\cdot(x+2)}=2^2 \\ \Leftrightarrow3\cdot(x+2)=2 \\ \Leftrightarrow3x=-4 \\ \Leftrightarrow x=-\frac{4}{3} \end{gathered}[/tex]
