The following exponential forms are equal to another one. Pick the right answer.
\left( \dfrac35 \right)^{2+x}
If a, b and c are three real numbers with c\gt0, then:
c^{a+b}=c^a\times c^b
Therefore:
\left( \dfrac35 \right)^{2+x} = \left( \dfrac35 \right)^{2} \left( \dfrac35 \right)^{x}
\left( \dfrac35 \right)^{2+x} =\left( \dfrac{3^2} {5^2} \right) \left( \dfrac35 \right)^{x}
\left( \dfrac35 \right)^{2+x} =\dfrac{9}{25} \left( \dfrac{3}{5} \right)^x
3^3 \times 3^{x}
If a, b and c are three real numbers with c\gt0, then:
c^a\times c^b=c^{a+b}
Therefore:
3^3 \times 3^x =3^{3+x}
a^5 \times a^{-2}
If a, b and c are three real numbers with c\gt0, then:
c^a\times c^b=c^{a+b}
Therefore:
a^5 \times a^{-2} =a^{5+\left(-2\right)}
a^5 \times a^{-2} =a^{5-2}
a^5 \times a^{-2} =a^{3}
2^{x+5}
If a, b and c are three real numbers with c\gt0, then:
c^{a+b}=c^a\times c^b
Therefore:
2^{x+5}=2^x \times 2^5
2^{x+5}=2^x \times 32
a^2 \times a^{3}
If a, b and c are three real numbers with c\gt0, then:
c^a\times c^b=c^{a+b}
Therefore:
a^2 \times a^{3}=a^{2+3}
a^2 \times a^{3}=a^5
6^{2+x}
If a, b and c are three real numbers with c\gt0, then:
c^{a+b}=c^a\times c^b
Therefore
6^{2+x} =6^2 \times 6^x
6^{2+x}=36 \times 6^x
2^{3+6}
If a, b and c are three real numbers with c\gt0, then:
c^{a+b} =c^a\times c^b
Therefore:
2^{3+6}=2^{3}\times 2^6