SELINA Solution Class 9 Chapter 7 Indices (Exponents) Exercise 7B

Question 1

Solve for x : 2^2x+1 = 8

Sol:

2^2x+1 = 8
⇒ 2^2x+1 = 2³
We know that if bases are equal, the powers are equal
⇒ 2x + 1 = 3
⇒ 2x = 3 - 1
⇒ 2x = 2
⇒ x = ${\displaystyle \frac{2}{2}}$
⇒ x = 1

Question 1.2

Solve for x : 2^5x-1 = 4 2^{3x + 1}

Sol:

 2^5x-1 = 4  x 2^{3x + 1}
⇒  2^5x-1 = 2² x 2^{3x + 1}
⇒  2^5x-1 = 2^{2 + 3x + 1}
⇒  2^5x-1 = 2^{3x + 3}
We know that if bases are equal, the powers are equal
⇒ 5x - 1 = 3x + 3
⇒ 5x - 3x = 3 + 1
⇒ 2x = 4
⇒ x = ${\displaystyle \frac{4}{2}}$
⇒ x = 2.

Question 1.3

Solve for x :  3^{4x + 1} = (27)^{x + 1}

Sol:

3^{4x + 1} = (27)^{x + 1}
⇒ 3^{4x + 1} = (3³)^{x + 1}
⇒ 3^{4x + 1} = 3^{3x + 1}
We know that if bases are equal, the powers are equal
⇒ 4x + 1 = 3x + 3
⇒ 4x - 3x = 3 - 1
⇒ x = 2.

Question 1.4

Solve for x : (49)^{x + 4} = 7² x (343)^{x + 1}

Sol:

(49)^{x + 4} = 7² x (343)^{x + 1}

⇒ ( 7 x 7 )^{x + 4} = 7² ( 7 x 7 x 7 )^{( x + 1 )}

⇒ ( 7² )^{x + 4}  = 7²( 7³ )^{( x + 1 )}

⇒ 7^{( 2x + 8 )} = 7² x 7^{3x + 3}

⇒ 7^{( 2x + 8 )} = 7^{3x + 3 + 2}

⇒ 7^{( 2x + 8 )} = 7^{3x + 5}
We know that if bases are equal, the powers are equal
⇒ 2x + 8 = 3x + 5
⇒ 3x - 2x = 8 - 5
⇒ x = 3

Question 2.1

Find x, if : 4^2x = ${\displaystyle \frac{1}{32}}$

Sol:

4^2x = ${\displaystyle \frac{1}{32}}$

⇒ ( 2 x 2 )^2x = ${\displaystyle \frac{1}{2\times 2\times 2\times 2\times 2}}$

⇒ ( 2² )^2x = ${\displaystyle \frac{1}{2^5}}$

⇒ 2^{2 x 2x} = 2^{- 5}

⇒ 2^4x = 2^{- 5}

We know that if bases are equal, the powers are equal
⇒ 4x = - 5
⇒ x = ${\displaystyle \frac{-5}{4}}$

Question 2.2

Find x, if : ${\displaystyle \sqrt{2^{x+3}}=16}$

Sol:

${\displaystyle \sqrt{2^{x+3}}=16}$

${\displaystyle {\left(2^{x+3}\right)}^{\frac{1}{2}}=2\times 2\times 2\times 2}$

⇒ ${\displaystyle {\left(2\right)}^{\frac{x+3}{2}}=2^4}$

We know that if bases are equal, the powers are equal.
⇒ ${\displaystyle \frac{x+3}{2}=4}$

⇒ x + 3 = 8
⇒ x = 8 - 3
⇒ x = 5

Question 2.3

Find x, if : ${\displaystyle {\left(\sqrt{\frac{3}{5}}\right)}^{x+1}=\frac{125}{27}}$

Sol:

${\displaystyle {\left(\sqrt{\frac{3}{5}}\right)}^{x+1}=\frac{125}{27}}$

⇒ ${\displaystyle {\left[{\left(\frac{3}{5}\right)}^{\frac{1}{2}}\right]}^{x+1}=\frac{5\times 5\times 5}{3\times 3\times 3}}$

⇒ ${\displaystyle {\left(\frac{3}{5}\right)}^{\frac{x+1}{2}}={\left(\frac{5}{3}\right)}^3}$

⇒ ${\displaystyle {\left(\frac{3}{5}\right)}^{\frac{x+1}{2}}={\left(\frac{3}{5}\right)}^{-}3}$

We know that if bases are equal, the powers are equal
⇒ ${\displaystyle \frac{x+1}{2}=-3}$

⇒ x + 1 = - 6
⇒ x = - 6 - 1
⇒ x = - 7

Question 2.4

Find x, if : ${\displaystyle {\left(\sqrt[3]{\frac{2}{3}}\right)}^{x-1}=\frac{27}{8}}$

Sol:

${\displaystyle {\left(\sqrt[3]{\frac{2}{3}}\right)}^{x-1}=\frac{27}{8}}$

${\displaystyle {\left[{\left(\frac{2}{3}\right)}^{\frac{1}{3}}\right]}^{x-1}=\frac{3^3}{2^3}}$

⇒ ${\displaystyle {\left(\frac{2}{3}\right)}^{\frac{x-1}{3}}={\left(\frac{3}{2}\right)}^3}$

⇒ ${\displaystyle {\left(\frac{2}{3}\right)}^{\frac{x-1}{3}}={\left(\frac{2}{3}\right)}^{-3}}$

We know that if bases are equal, the powers are equal
⇒ ${\displaystyle \frac{x-1}{3}=-3}$

⇒ x - 1 = - 9
⇒ x = - 9 + 1
⇒ x = - 8

Question 3.1

Solve :  4^{x - 2} - 2^{x + 1} = 0

Sol:

4^{x - 2} - 2^{x + 1} = 0
⇒ 4^{x - 2} = 2^{x + 1}
⇒ (2²)^{x - 2}  = 2^{x + 1}
⇒ 2^{2x - 4}  = 2^{x + 1}
We know that if bases are equal, the powers are equal
⇒ 2x - 4 = x + 1
⇒ 2x - x = 4 + 1
⇒ x = 5.

Question 3.2

Solve : ${\displaystyle {\left[3^x\right]}^2}$ : 3x = 9 : 1

Sol:

${\displaystyle {\left[3^x\right]}^2}$ : 3x = 9 : 1

⇒ ${\displaystyle \frac{{\left[3^x\right]}^2}{3^x}=\frac{9}{1}}$

⇒ ${\displaystyle {\left[3^x\right]}^2=9\times 3^x}$

⇒ ${\displaystyle {\left[3^x\right]}^2=3^2\times 3^x}$

⇒ ${\displaystyle {\left[3^x\right]}^2=3^{x+2}}$

We know that if bases are equal, the powers are equal.
⇒ x² = x + 2
⇒ x² - x - 2 = 0
⇒ x² - 2x + x - 2 = 0
⇒ x( x - 2 ) + 1( x - 2 ) = 0
⇒ ( x + 1 )( x - 2 ) = 0
⇒ x + 1 = 0            or      x - 2 = 0
⇒ x = - 1                or      x = 2.

Question 4.1

Solve : 8 x 2^2x + 4 x 2^{x + 1} = 1 + 2^x

Sol:

8 x 2^2x + 4 x 2^{x + 1} = 1 + 2^x

⇒ 8 x ${\displaystyle {\left(2^x\right)}^2}$ + 4 x 2^x x 2¹ = 1 + 2^x
⇒ 8 x ${\displaystyle {\left(2^x\right)}^2}$ + 4 x 2^x x 2¹ - 1 - 2^x = 0
⇒ 8 x ${\displaystyle {\left(2^x\right)}^2}$ + 2^x  x ( 8 - 1 ) - 1 = 0
⇒ 8 x ${\displaystyle {\left(2^x\right)}^2}$ + 7( 2^x  ) - 1 = 0
⇒ 8y² + 7y - 1 = 0                   [ y = 2^x ]
⇒ 8y² + 8y - y - 1 = 0
⇒ 8y( y + 1 ) - 1( y + 1 ) = 0
⇒ ( 8y - 1 )( y + 1 ) = 0

⇒ 8y = 1  or  y = - 1

⇒ y = ${\displaystyle \frac{1}{8}}$  or  y = -1

⇒ 2^x = ${\displaystyle \frac{1}{8}}$ or 2^x = - 1

⇒ 2^x = ${\displaystyle \frac{1}{2^3}}$ or 2^x = - 1

⇒ 2^x = ${\displaystyle 2^{-3}}$ or 2^x = - 1
⇒ x = - 3
[ ∵ 2^x  = - 1 is not possible. ]

Question 4.2

Solve : 2^2x + 2^x+2 - 4 x 2³ = 0

Sol:

2^2x + 2^x+2 - 4 x 2³ = 0
⇒ ( 2^x)² + 2^x. 2² - 4 x 2 x 2 x 2 = 0
⇒ ( 2^x)² + 2^x. 2² - 32  = 0
Putting y = 2^x
⇒ y² + 4y - 32 = 0
⇒ y2 + 8y - 4y - 32 = 0
⇒ y( y + 8 ) - 4( y + 8 ) = 0
⇒ ( y + 8 )( y - 4 ) = 0
⇒ y + 8 = 0  or  y - 4 = 0
⇒ y = - 8 or y = 4
⇒ 2^x = - 8 or 2^x = 4
⇒ 2^x = 2²
[ ∵ 2^x = - 8 is not possible. ]
⇒ x = 2.

Question 4.3

Solve : ${\displaystyle {\left(\sqrt{3}\right)}^{x-3}={\left(\sqrt[4]{3}\right)}^{x+1}}$

Sol:

${\displaystyle {\left(\sqrt{3}\right)}^{x-3}={\left(\sqrt[4]{3}\right)}^{x+1}}$

⇒ ${\displaystyle {\left(3^{\frac{1}{2}}\right)}^{x-3}={\left(3^{\frac{1}{4}}\right)}^{x+1}}$

⇒ ${\displaystyle 3^{\frac{x-3}{2}}=3^{\frac{x+1}{4}}}$

⇒ ${\displaystyle \frac{x-3}{2}=\frac{x+1}{4}}$

⇒ 4( x - 3 ) = 2( x + 1 )
⇒ 4x - 12 = 2x + 2
⇒ 4x - 2x = 12 + 2
⇒ 2x = 14
⇒ x = ${\displaystyle \frac{14}{2}}$
⇒ x = 7

Question 5

Find the values of m and n if : 
${\displaystyle 4^{2m}={\left(\sqrt[3]{16}\right)}^{-\frac{6}{n}}={\left(\sqrt{8}\right)}^2}$

Sol:

${\displaystyle 4^{2m}={\left(\sqrt[3]{16}\right)}^{-\frac{6}{n}}={\left(\sqrt{8}\right)}^2}$
⇒ ${\displaystyle 4^{2m}={\left(\sqrt{8}\right)}^2}$                    ....(1)

and
${\displaystyle {\left(\sqrt[3]{16}\right)}^{-\frac{6}{n}}={\left(\sqrt{8}\right)}^2}$     ....(2)
From (1)
${\displaystyle 4^{2m}={\left(\sqrt{8}\right)}^2}$

⇒ ${\displaystyle {\left(2^2\right)}^{2m}={\left(\sqrt{2^3}\right)}^2}$

⇒ ${\displaystyle 2^{4m}={\left[{\left(2^3\right)}^{\frac{1}{2}}\right]}^2}$

⇒ ${\displaystyle 2^{4m}={\left[2^{3\times \frac{1}{2}}\right]}^2}$

⇒ ${\displaystyle 2^{4m}= 2^{3\times \frac{1}{2}\times 2}}$

⇒ ${\displaystyle 2^{4m}=2^3}$

⇒ 4m = 3

⇒ m = ${\displaystyle \frac{3}{4}}$

From (2), We have
${\displaystyle {\left(3\sqrt{16}\right)}^{-\frac{6}{x}}={\left(\sqrt{8}\right)}^2}$

⇒ ${\displaystyle {\left(\sqrt[3]{2\times 2\times 2\times 2}\right)}^{-\frac{6}{x}}={\left(\sqrt{2\times 2\times 2}\right)}^2}$

⇒ ${\displaystyle {\left(\sqrt[3]{2^4}\right)}^{-\frac{6}{x}}={\left(\sqrt{2^3}\right)}^2}$

⇒ ${\displaystyle {\left[{\left(2^4\right)}^{\frac{1}{3}}\right]}^{-\frac{6}{x}}={\left[{\left(2^3\right)}^{\frac{1}{2}}\right]}^2}$

⇒ ${\displaystyle {\left[2^{\frac{4}{3}}\right]}^{-\frac{6}{x}}={\left[2^{\frac{3}{2}}\right]}^2}$

⇒ ${\displaystyle 2^{\frac{4}{3}\times (-\frac{6}{x})=2^{\frac{3}{2}\times 2}}}$

⇒ ${\displaystyle 2^{-\frac{8}{x}}=2^3}$

⇒ ${\displaystyle -\frac{8}{x}=3}$

⇒ ${\displaystyle x=-\frac{8}{3}\text{Thus m}=\frac{3}{4}x=-\frac{8}{3}}$

Question 6

Solve x and y if : ( √32 )^x ÷ 2^{y + 1} = 1 and 8^y - 16^{4 - x/2} = 0

Sol:

Consider the quation
${\displaystyle {\left(\sqrt{32}\right)}^x÷2^{y+1}=1}$
⇒ ${\displaystyle {\left(\sqrt{2\times 2\times 2\times 2\times 2}\right)}^x÷2^{y+1}=1}$
⇒ ${\displaystyle {\left(\sqrt{2^5}\right)}^x÷2^{y+1}=1}$
⇒ ${\displaystyle {\left[{\left(2^5\right)}^{\frac{1}{2}}\right]}^x÷2^{y+1}=x^0}$

⇒ ${\displaystyle 2^{5\frac{x}{2}}÷2^{y+1}=x^0}$

⇒ ${\displaystyle \frac{5x}{2}-(y+1)=0}$
⇒ 5x - 2( y + 1 ) = 0
⇒ 5x - 2y - 2 = 0                      ....(1)

Now consider the other equation
${\displaystyle 8^y-{16}^{4-\frac{x}{2}}=0}$
⇒ ${\displaystyle {\left(2^3\right)}^y-{\left(2^4\right)}^{4-\frac{x}{2}}=0}$

⇒ ${\displaystyle 2^{3y}-2^{4(4-\frac{x}{2})}=0}$

⇒ ${\displaystyle 2^{3y}=2^{4(4-\frac{x}{2})}}$

⇒ ${\displaystyle 3y=4(4-\frac{x}{2})}$

⇒ 3y = 16 - 2x
⇒ 2x + 3y = 16                      ...(2)

Thus, We have two equations,
5x - 2y = 2                           ...(1)
2x + 3y = 16                        ....(2)
Multiplying equation (1) by 3 and (2) by 2, We have
15x - 6y = 6                         ....(3)
4x + 6y = 32                        ....(4)
Adding equation (3) and (4), We have
19x = 38
⇒ x = 2 
Substituting the value of x in equation (1), We have
5(2) - 2y = 2
⇒ 10 - 2y = 2
⇒ 2y = 10 - 2
⇒ 2y = 8
⇒ y = ${\displaystyle \frac{8}{2}}$
⇒ y = 4
Thus the values of x and y are : x = 2 and y = 4.

Question 7.1

Prove that : ${\displaystyle {\left(\frac{x^a}{x^b}\right)}^{a+b-c}{\left(\frac{x^b}{x^c}\right)}^{b+c-a}{\left(\frac{x^c}{x^a}\right)}^{c+a-b}}$

Sol:

LHS = ${\displaystyle {\left(\frac{x^a}{x^b}\right)}^{a+b-c}{\left(\frac{x^b}{x^c}\right)}^{b+c-a}{\left(\frac{x^c}{x^a}\right)}^{c+a-b}}$

= ${\displaystyle {\left(x^{a-b}\right)}^{a+b-c}\times {\left(x^{b-c}\right)}^{b+c-a}\times {\left(x^{c-a}\right)}^{c+a-b}}$

= ${\displaystyle x^{(a-b)(a+b-c)}\times x^{(b-c)(b+c-a)}\times x^{(c-a)(c+a-b)}}$

= ${\displaystyle x^{a^2+ab-ac-ab-b^2+bc}\times x^{b^2+bc-ab-cd-c^2+ac}\times x^{c^2+ac-bc-ac-a^2+ab}}$

= ${\displaystyle x^{a^2-ac-b^2+bc+b^2-ab-c^2+ac+c^2-bc-a^2+ab}}$
= x⁰
= 1
= RHS

Question 7.2

Prove that :
${\displaystyle \frac{x^{a(b-c)}}{x^b(a-c)}÷{\left(\frac{x^b}{x^a}\right)}^c=1}$

Sol:

We need to prove that
${\displaystyle \frac{x^{a(b-c)}}{x^b(a-c)}÷{\left(\frac{x^b}{x^a}\right)}^c=1}$

LHS =

= ${\displaystyle x^{a(b-c)-b(a-c)}÷\frac{x^{bc}}{x^{ac}}}$ 

= ${\displaystyle x^{ab-ac-ab+bc}÷x^{bc-ac}}$

= ${\displaystyle x^{ab-ac-ab+bc-bc+ac}}$
= ${\displaystyle x^0}$
= 1
= RHS

Question 8

If a^x = b, b^y = c and c^z = a, prove that : xyz = 1.

Sol:

We are given that
a^x = b, b^y = c and c^z = a
Consider the equation
a^x = b
⇒ a^xyz = b^yz      [ raising to the power yz on both sides ] 
⇒  a^xyz = (b^y)^z
⇒ a^xyz = c^z         [ ∵ b^y = c ]
⇒  a^xyz = c^z
⇒  a^xyz = a         [ ∵ c^z = a ]
⇒  a^xyz = a¹
⇒  xyz = 1

Question 9

If a^x = b^y = c^z and b² = ac, prove that : y = ${\displaystyle \frac{2az}{x+z}}$

Sol:

Let ax = by = cz = k
∴ a = ${\displaystyle k^{\frac{1}{x}};b=k^{\frac{1}{y}};c=k^{\frac{1}{z}}}$

Also, We have b2 = ac 
∴ ${\displaystyle {\left(k^{\frac{1}{y}}\right)}^2=\left(k^{\frac{1}{x}}\right)\times \left(k^{\frac{1}{2}}\right)}$ 

⇒ ${\displaystyle k^{\frac{2}{y}}=k^{\frac{1}{x}+\frac{1}{z}}}$

⇒ ${\displaystyle k^{\frac{2}{y}}=\frac{k^{z+x}}{ xz}}$

Comparing the powers we have
${\displaystyle \frac{2}{y}=\frac{z+x}{xz}}$

⇒ ${\displaystyle y=\frac{2xz}{z+x}}$

Question 10

If 5^-P = 4^-q = 20^r, show that : ${\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=0}$

Sol:

Let 5^-P = 4^-q = 20^r = k
5^-P = k ⇒ 5 = ${\displaystyle k^{-\frac{1}{p}} [\because a^p=b^q\Rightarrow a=b^{\frac{q}{p}}]}$

4^-q = k ⇒ 4 = ${\displaystyle k^{-\frac{1}{q}} [\because a^p=b^q\Rightarrow a=b^{\frac{q}{p}}]}$

20^r = k ⇒ 20 = ${\displaystyle k^{\frac{1}{r}} [\because a^p=b^q\Rightarrow a=b^{\frac{q}{p}}]}$

5 x 4 = 20
⇒ ${\displaystyle k^{-\frac{1}{p}}\times k^{-\frac{1}{q}}=k^{\frac{1}{r}}}$ 

⇒ ${\displaystyle k^{-\frac{1}{p}-\frac{1}{q}}=k^{\frac{1}{r}}}$

⇒ ${\displaystyle k^0=k^{\frac{1}{p}+\frac{1}{q}+\frac{1}{r}}}$

If bases are equal, powers are also equal.
⇒  ${\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=0}$

Question 11

If m ≠ n and (m + n)^-1 (m^-1 + n^-1) = m^xn^y, show that : x + y + 2 = 0

Sol:

(m + n)^-1 (m^-1 + n^-1) = m^xn^y

⇒ ${\displaystyle \frac{1}{m+n}\times (\frac{1}{m}+\frac{1}{n})=m^x.n^y}$

⇒ ${\displaystyle \frac{1}{m+n}\times \left(\frac{m+n}{mn}\right)=m^x.n^y}$

⇒ ${\displaystyle \frac{1}{mn}=m^x.n^y}$

⇒ ${\displaystyle m^{-1}{n}^{-1}=m^x.n^y}$
Comparing the coefficient of x and y, we get
x = - 1 and y = -1
LHS
y + y + 2 = ( - 1) + ( - 1) + 2 = 0 = RHS

Question 12

If 5^{x + 1} = 25^{x - 2}, find the value of  3^{x - 3} × 2^{3 - x}.

Sol:

5^{x + 1} = 25^{x - 2} 
⇒ 5^{x + 1} = (5²)^{x - 2}
⇒  5^{x + 1} = 5^{2x - 4}
If bases are equal, powers are also equal.
⇒ x + 1 = 2x - 4
⇒ 2x - x = 4 + 1
⇒ x = 5

∴ 3^{x - 3} x 2^{3 - x} 
= 3^{5 - 3} x 2^{3 - 5}
= 3² x 2^-2 
= 9 x ${\displaystyle \frac{1}{4}=\frac{9}{4}}$

Question 13

If 4^{x + 3} = 112 + 8 × 4^x, find the value of (18x)^3x.

Sol:

4^{x + 3} = 112 + 8 × 4^x 
⇒ 4^x x 4³ = 112 + 8 x 4^x 
⇒  64 x 4^x = 112 + 8 x 4^x 
Let 4^x = y
64y = 112 + 8y
⇒ 56y = 112
⇒ y = 2
Substituting We get,
4^x = 2
⇒ 2^2x = 2
⇒ 2x = 1
⇒ x = ${\displaystyle \frac{1}{2}}$

${\displaystyle {\left(18x\right)}^{3x} ={\left(\frac{18}{2}\right)}^{3\times \frac{1}{2}}=9^{3\times \frac{1}{2}}={\left(9^{\frac{1}{2}}\right)}^3=3^3=27}$

Question 14.1

Solve for x :  4^x-1 × (0.5)^{3 - 2x} = ${\displaystyle {\left(\frac{1}{8}\right)}^{-x}}$

SoL:

4^x-1 × (0.5)^{3 - 2x} = ${\displaystyle {\left(\frac{1}{8}\right)}^{-x}}$

⇒ ${\displaystyle {\left(2^2\right)}^{x-1}\times {\left(\frac{1}{2}\right)}^{3-2x}={\left(\frac{1}{2^3}\right)}^{-x}}$

⇒ 2^{2x - 2} x 2^{-( 3 - 2x )} = (2 ^-3)^{- x}

⇒ 2^{2x - 2 - 3 + 2x} = 2^3x

⇒ 2^{4x - 5} = 2^3x
⇒ 4x - 5 = 3x
⇒ 4x - 3x = 5
⇒ x = 5

Question 14.2

Solve for x :  (a^{3x + 5})². (a^x)⁴ = a^{8x + 12}

Sol:

(a^{3x + 5})². (a^x)⁴ = a^{8x + 12} 

⇒ a^{6x + 10 + 4x =} a^{8x + 12}
If bases are the same, powers are also same
⇒ 10x + 10 = 8x + 12
⇒ 2x = 2
⇒ x = 1

Question 14.3

Solve for x : ${\displaystyle {\left(81\right)}^{\frac{3}{4}}-{\left(\frac{1}{32}\right)}^{-\frac{2}{5}}+x{\left(\frac{1}{2}\right)}^{-1}{.2}^0=27}$

Sol:

${\displaystyle {\left(81\right)}^{\frac{3}{4}}-{\left(\frac{1}{32}\right)}^{-\frac{2}{5}}+x{\left(\frac{1}{2}\right)}^{-1}{.2}^0=27}$

⇒ ${\displaystyle 3^{4\times \frac{3}{4}}-{\left(2^{-}5\right)}^{-\frac{2}{5}}+x\left(2\right)=27}$

⇒ ${\displaystyle 3^3-2^2+2x=27}$
⇒ 2x + 27 - 4 = 27
⇒ 2x = 4
⇒ x = 2

Question 14.4

Solve for x : 2^{3x + 3} = 2^{3x + 1} + 48

Sol:

2^{3x + 3} = 2^{3x + 1} + 48
⇒ 8 x 2^3x = 2^3x x 2 + 48
⇒ 2^3x ( 8 - 2 ) = 48
⇒ 2^3x x 6  = 48
⇒ 2^3x = 8
⇒ 2^3x = 2³
⇒ 3x = 3
⇒ x = 1

Question 14.5

Solve for x :  3(2^x + 1) - 2^{x + 2} + 5 = 0

Sol:

3(2^x + 1) - 2^{x + 2} + 5 = 0
⇒ 3 x 2^x + 3 - 2^x x 2² + 5 = 0
⇒ 2^x ( 3 - 4 ) + 8 = 0
⇒ - 2^x = - 8
⇒ 2^x = 2³
⇒ x = 3

Question 14.6

Solve for x : 9^x+2 = 720 + 9^x

Sol:

9^x+2 = 720 + 9^x
⇒ 9^x+2 - 9^x = 720
⇒ 9^x (9² - 1) = 720
⇒ 9^x (81 - 1) = 720
⇒ 9^x (80) = 720
⇒ 9^x = 9
⇒ 9^x = 9¹
⇒ x = 1