Solution:

\(2^{x+2}\times8^x=1,\)

\(\Rightarrow 2^{x+2}\times2^{3x}=2^0\) HintExpressing the equation to common bases

\(\Rightarrow(x+2)+3x=0\) HintSince the bases are the same, we equate the exponents

\(\Rightarrow4x=-2\) HintDeviding both side by 4

\(\Rightarrow x=-\frac{1}{2}.\)

By:  Asamoah

Solution:

\(\sqrt{\Big(\frac{x^3y^5}{xy^7}\Big)}\)

\(=\Big(\frac{x^3y^5}{xy^7}\Big)^{\frac{1}{2}}\) HintRemoving the square root by applying the “fractional index rule”

\(=\Big( \frac{x^2}{y^2}\Big)^{\frac{1}{2}}\)

\(=\frac{x^{2\times\frac{1}{2}}}{y^{2\times\frac{1}{2}}}\)

\(=\frac{x^1}{y^1}\)

\(=\frac{x}{y}.\)

By:  Asamoah

Solution:

\(\frac{64^n\times2}{16^{1-n}}=4^{2n}.\)

\(\Rightarrow \frac{2^{6(n)}\times2^1}{2^{4(1-n)}}=2^{2(2n)}\) HintExpressing the equation to common bases

\(\Rightarrow 2^{6(n)}\times2^1=2^{4(1-n)}\times2^{2(2n)}\) HintCross-multiplication method applied

\(\Rightarrow 6n+1=4(1-n)+2(2n)\) HintSince the bases are the same, we equate the exponents

\(\Rightarrow 6n+1=4-4n-4n\)

\(\Rightarrow 6n+4n-4n=4-1\)

\(\Rightarrow 6n=3\) HintDeviding both side by 6

\(\Rightarrow n=\frac{1}{2}.\)

By:  Asamoah

Solution:

\(4^{2-x}\times16^{x+1}=64.\)

\(\Rightarrow 4^{2-x}\times4^{2(x+1)}=4^3\) HintEquation expressed to common bases

\(\Rightarrow2-x+2(x+1)=3\) HintSince the bases are the same, we equate the exponents

\(\Rightarrow 2-x+2x+2=3\)

\(\Rightarrow -x+2x=3-2-2\)

\(\Rightarrow x=-1.\)

By:  Asamoah