WASSCE Indices is one of the topics in WASSCE mathematics syllabus. The followings are WASSCE past questions and solutions on Indices. Candidates are advised to try each question by themselves before viewing solution. Frequent practice guarantees success in exams. VISIT MATHS LIBRARY FOR MORE TOPICS

WASSCE 2013 private candidates. If \(2^{x+2}\times8^x=1,\) find the value of \(x.\)

Solution Solution:

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

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

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

\(\Rightarrow4x=-2\) Hint Deviding both side by 4

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

By: Asamoah

WASSCE 2010 private candidates. Simplify \(\sqrt{\Big(\frac{x^3y^5}{xy^7}\Big)},\) where \(x>0\) and \(y>0.\)

Solution Solution:

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

\(=\Big(\frac{x^3y^5}{xy^7}\Big)^{\frac{1}{2}}\) Hint Removing 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

WASSCE 2009 school candidates. Solve the equation \(\frac{64^n\times2}{16^{1-n}}=4^{2n}.\)

Solution Solution:

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

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

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

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

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

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

\(\Rightarrow 6n=3\) Hint Deviding both side by 6

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

By: Asamoah

WASSCE 2006 school candidates. Solve the equation \(4^{2-x}\times16^{x+1}=64.\)

Solution Solution:

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

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

\(\Rightarrow2-x+2(x+1)=3\) Hint Since 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

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Thanks can I have more passed questions and if possible answers

Please visit the following links:

BECE https://www.mywschool.com/waec-bece-past-questions/

WASSCE https://www.mywschool.com/waec-sssce-wassce-past-questions/

can I have all waec past questions from 1998 to 2018

Check here: https://www.mywschool.com/past-questions/