Indices (Exponents) — ICSE Class 9 Maths Important Questions

Answer: (a) $\dfrac18$18.

$2^{-3}=\dfrac{1}{2^3}=\dfrac18$2^-3=1/2^3=18.

Answer: (a) $x^3$x^3.

$x^5\times x^{-2}=x^{5+(-2)}=x^3$x^5× x^-2=x^5+(-2)=x^3.

Answer: (a) $\dfrac49$49.

$\left(\dfrac{27}{8}\right)^{-2/3}=\left(\dfrac{8}{27}\right)^{2/3}=\left(\dfrac{2}{3}\right)^{2}=\dfrac49$(27/8)^-2/3=(8/27)^2/3=(2/3)^2=49.

Answer: (a) $27$27.

$2^x=2^{3(y+1)}\Rightarrow x=3y+3$2^x=2^3(y+1) x=3y+3. Also $3^{2y}=3^{x-9}\Rightarrow 2y=x-9$3^2y=3^x-9 2y=x-9. Substitute: $2y=(3y+3)-9\Rightarrow -y=-6\Rightarrow y=6$2y=(3y+3)-9 -y=-6 y=6, so $x=21$x=21. Hence $x+y=27$x+y=27.

Answer: (a) Using R, $\left(\dfrac{3}{5}\right)^{-2}=\left(\dfrac{5}{3}\right)^{2}=\dfrac{25}{9}$(3/5)^-2=(5/3)^2=25/9, so A is true and R correctly explains it.

$\left(\dfrac{16}{81}\right)^{-3/4}=\left(\dfrac{81}{16}\right)^{3/4}=\left(\dfrac{3^4}{2^4}\right)^{3/4}=\left(\dfrac{3}{2}\right)^{3}=\dfrac{27}{8}.$(16/81)^-3/4=(81/16)^3/4=(3^4/2^4)^3/4=(3/2)^3=27/8.

Write $27=3^3$27=3^3:

$3^{x+2}=3^{3}\Rightarrow x+2=3\Rightarrow x=1.$3^x+2=3^3 x+2=3 x=1.

Write with positive indices:

$\frac{\dfrac1a+\dfrac1b}{\dfrac1a-\dfrac1b}=\frac{\dfrac{b+a}{ab}}{\dfrac{b-a}{ab}}.$1a+1b/1a-1b=b+a/abb-a/ab.

The $ab$ab cancels:

$=\frac{a+b}{b-a}.$=a+b/b-a.

Consider the second term and multiply numerator and denominator by $x^{a-b}$x^a-b:

$\frac{1}{1+x^{b-a}}=\frac{x^{a-b}}{x^{a-b}(1+x^{b-a})}=\frac{x^{a-b}}{x^{a-b}+x^{0}}=\frac{x^{a-b}}{x^{a-b}+1}.$11+x^b-a=x^a-bx^a-b(1+x^b-a)=x^a-bx^a-b+x^0=x^a-bx^a-b+1.

Now add the two terms:

$\frac{1}{1+x^{a-b}}+\frac{x^{a-b}}{1+x^{a-b}}=\frac{1+x^{a-b}}{1+x^{a-b}}=1.$11+x^a-b+x^a-b1+x^a-b=1+x^a-b1+x^a-b=1.

Hence proved.

Express every term as a power of $2$2:

$4^2=(2^2)^2=2^4,\qquad 8^3=(2^3)^3=2^9.$4^2=(2^2)^2=2^4, 8^3=(2^3)^3=2^9.

So

$\frac{2^5\times 2^4}{2^9}=\frac{2^{9}}{2^{9}}=2^{0}=1.$2^5× 2^4/2^9=2^92^9=2^0=1.

Part 1: Write $\dfrac18=2^{-3}$18=2^-3:

$2^{2x-1}=2^{-3}\Rightarrow 2x-1=-3\Rightarrow 2x=-2\Rightarrow x=-1.$2^2x-1=2^-3 2x-1=-3 2x=-2 x=-1.

Part 2: Write $25=5^2$25=5^2 so $25^{x-1}=5^{2(x-1)}=5^{2x-2}$25^x-1=5^2(x-1)=5^2x-2:

$5^{x+1}=5^{2x-2}\Rightarrow x+1=2x-2\Rightarrow x=3.$5^x+1=5^2x-2 x+1=2x-2 x=3.

So the solutions are $x=-1$x=-1 and $x=3$x=3 respectively.

Let $a^x=b^y=c^z=k$a^x=b^y=c^z=k (a common value, $k>0$k>0).

Then

$a=k^{1/x},\qquad b=k^{1/y},\qquad c=k^{1/z}.$a=k^1/x, b=k^1/y, c=k^1/z.

Given $b^2=ac$b^2=ac, substitute:

$\left(k^{1/y}\right)^2=k^{1/x}\cdot k^{1/z}.$(k^1/y)^2=k^1/x· k^1/z.

$k^{2/y}=k^{\,\frac1x+\frac1z}.$k^2/y=k^\,1x+1z.

Since the bases are equal ($k>0,\ k\neq1$k>0, k≠1), equate the exponents:

$\frac2y=\frac1x+\frac1z.$2y=1x+1z.

Hence $\dfrac1x+\dfrac1z=\dfrac2y$1x+1z=2y, as required.

(i) $N=500\cdot 2^{3}=500\times8=4000$N=500· 2^3=500×8=4000 bacteria.

(ii) Solve $500\cdot 2^{t}=16000\Rightarrow 2^{t}=32=2^{5}\Rightarrow t=5$500· 2^t=16000 2^t=32=2^5 t=5 hours.

(iii) $500=2^2\times5^3$500=2^2×5^3. The $3$3-hour count $4000=2^{5}\times5^{3}$4000=2^5×5^3 (since $4000=4000$4000=4000, and $2^5=32,\ 32\times125=4000$2^5=32, 32×125=4000).

(iv) With tripling, $N=500\cdot 3^{t}$N=500· 3^t.