What is the general solution for 9^{sin(x)-1}+3=3^{sin(x)}+3^{sin(x)-1} ?

The general solution for 9^{sin(x)-1}+3=3^{sin(x)}+3^{sin(x)-1} is x= pi/2+2pin

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9^{sin(x)-1}+3=3^{sin(x)}+3^{sin(x)-1}

Solution

9^{s i n (x) - 1} + 3 = 3^{s i n (x)} + 3^{s i n (x) - 1}

Solution

x = \frac{π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

9^{s i n (x) - 1} + 3 = 3^{s i n (x)} + 3^{s i n (x) - 1}

Solve by substitution

9^{s i n (x) - 1} + 3 = 3^{s i n (x)} + 3^{s i n (x) - 1}

Let:

sin (x) = u

9^{u - 1} + 3 = 3^{u} + 3^{u - 1}

9^{u - 1} + 3 = 3^{u} + 3^{u - 1} : u = 2, u = 1

9^{u - 1} + 3 = 3^{u} + 3^{u - 1}

Simplify

3^{u} + 3^{u - 1} : 4 \cdot 3^{u - 1}

3^{u} + 3^{u - 1}

3^{u} = 3^{u - 1 + 1}

= 3^{u - 1 + 1} + 3^{u - 1}

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

3^{u - 1 + 1} = 3^{1} \cdot 3^{u - 1}

= 3^{1} \cdot 3^{u - 1} + 3^{u - 1}

Factor out common term

3^{u - 1}

= 3^{u - 1} (3^{1} + 1)

Refine

= 4 \cdot 3^{u - 1}

9^{u - 1} + 3 = 4 \cdot 3^{u - 1}

Apply exponent rules

9^{u - 1} + 3 = 4 \cdot 3^{u - 1}

Convert to base

3 : 3^{2 (u - 1)} + 3 = 4 \cdot 3^{u - 1}

Convert

9

to base

3

9 = 3^{2}

(3^{2})^{u - 1} + 3 = 4 \cdot 3^{u - 1}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

(3^{2})^{u - 1} = 3^{2 (u - 1)}

3^{2 (u - 1)} + 3 = 4 \cdot 3^{u - 1}

3^{2 (u - 1)} + 3 = 4 \cdot 3^{u - 1}

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

3^{2 (u - 1)} = 3^{2 u} \cdot 3^{- 2}, 3^{u - 1} = 3^{u} \cdot 3^{- 1}

3^{2 u} \cdot 3^{- 2} + 3 = 4 \cdot 3^{u} \cdot 3^{- 1}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

3^{2 u} = (3^{u})^{2}

(3^{u})^{2} \cdot 3^{- 2} + 3 = 4 \cdot 3^{u} \cdot 3^{- 1}

(3^{u})^{2} \cdot 3^{- 2} + 3 = 4 \cdot 3^{u} \cdot 3^{- 1}

Rewrite the equation with

3^{u} = v

(v)^{2} \cdot 3^{- 2} + 3 = 4 v \cdot 3^{- 1}

Solve

v^{2} \cdot 3^{- 2} + 3 = 4 v \cdot 3^{- 1} : v = 9, v = 3

v^{2} \cdot 3^{- 2} + 3 = 4 v \cdot 3^{- 1}

Expand

v^{2} \cdot 3^{- 2} + 3 : \frac{1}{9} v^{2} + 3

v^{2} \cdot 3^{- 2} + 3

3^{- 2} = \frac{1}{9}

3^{- 2}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{3 ^{2}}

3^{2} = 9

= \frac{1}{9}

= \frac{1}{9} v^{2} + 3

Expand

4 v \cdot 3^{- 1} : \frac{4}{3} v

4 v \cdot 3^{- 1}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

3^{- 1} = \frac{1}{3}

= 4 \cdot \frac{1}{3} v

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 4}{3} v

Multiply the numbers:

1 \cdot 4 = 4

= \frac{4}{3} v

\frac{1}{9} v^{2} + 3 = \frac{4}{3} v

Move

\frac{4}{3} v

to the left side

\frac{1}{9} v^{2} + 3 = \frac{4}{3} v

Subtract

\frac{4}{3} v

from both sides

\frac{1}{9} v^{2} + 3 - \frac{4}{3} v = \frac{4}{3} v - \frac{4}{3} v

Simplify

\frac{1}{9} v^{2} + 3 - \frac{4}{3} v = 0

\frac{1}{9} v^{2} + 3 - \frac{4}{3} v = 0

Write in the standard form

a x^{2} + b x + c = 0

\frac{v ^{2}}{9} - \frac{4 v}{3} + 3 = 0

Solve with the quadratic formula

\frac{v ^{2}}{9} - \frac{4 v}{3} + 3 = 0

Quadratic Equation Formula:

For

a = \frac{1}{9}, b = - \frac{4}{3}, c = 3

v_{1, 2} = \frac{- ( - \frac{4}{3} ) \pm ( - \frac{4}{3} ) ^{2} - 4 \cdot \frac{1}{9} \cdot 3}{2 \cdot \frac{1}{9}}

v_{1, 2} = \frac{- ( - \frac{4}{3} ) \pm ( - \frac{4}{3} ) ^{2} - 4 \cdot \frac{1}{9} \cdot 3}{2 \cdot \frac{1}{9}}

(- \frac{4}{3})^{2} - 4 \cdot \frac{1}{9} \cdot 3 = \frac{2}{3}

(- \frac{4}{3})^{2} - 4 \cdot \frac{1}{9} \cdot 3

(- \frac{4}{3})^{2} = \frac{4 ^{2}}{3 ^{2}}

(- \frac{4}{3})^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- \frac{4}{3})^{2} = (\frac{4}{3})^{2}

= (\frac{4}{3})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{4 ^{2}}{3 ^{2}}

4 \cdot \frac{1}{9} \cdot 3 = \frac{4}{3}

4 \cdot \frac{1}{9} \cdot 3

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 4 \cdot 3}{9}

Multiply the numbers:

1 \cdot 4 \cdot 3 = 12

= \frac{12}{9}

Cancel the common factor:

3

= \frac{4}{3}

= \frac{4 ^{2}}{3 ^{2}} - \frac{4}{3}

\frac{4 ^{2}}{3 ^{2}} = \frac{16}{9}

\frac{4 ^{2}}{3 ^{2}}

4^{2} = 16

= \frac{16}{3 ^{2}}

3^{2} = 9

= \frac{16}{9}

= \frac{16}{9} - \frac{4}{3}

Join

\frac{16}{9} - \frac{4}{3} : \frac{4}{9}

\frac{16}{9} - \frac{4}{3}

Least Common Multiplier of

9, 3 : 9

9, 3

Least Common Multiplier (LCM)

Prime factorization of

9 : 3 \cdot 3

9

9

divides by

39 = 3 \cdot 3

= 3 \cdot 3

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Multiply each factor the greatest number of times it occurs in either

9

3

= 3 \cdot 3

Multiply the numbers:

3 \cdot 3 = 9

= 9

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

9

For

\frac{4}{3} :

multiply the denominator and numerator by

3

\frac{4}{3} = \frac{4 \cdot 3}{3 \cdot 3} = \frac{12}{9}

= \frac{16}{9} - \frac{12}{9}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{16 - 12}{9}

Subtract the numbers:

16 - 12 = 4

= \frac{4}{9}

= \frac{4}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{4}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{4}{3}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2}{3}

v_{1, 2} = \frac{- ( - \frac{4}{3} ) \pm \frac{2}{3}}{2 \cdot \frac{1}{9}}

Separate the solutions

v_{1} = \frac{- ( - \frac{4}{3} ) + \frac{2}{3}}{2 \cdot \frac{1}{9}}, v_{2} = \frac{- ( - \frac{4}{3} ) - \frac{2}{3}}{2 \cdot \frac{1}{9}}

v = \frac{- ( - \frac{4}{3} ) + \frac{2}{3}}{2 \frac{1}{9}} : 9

\frac{- ( - \frac{4}{3} ) + \frac{2}{3}}{2 \cdot \frac{1}{9}}

Apply rule

- (- a) = a

= \frac{\frac{4}{3} + \frac{2}{3}}{2 \cdot \frac{1}{9}}

Multiply

2 \cdot \frac{1}{9} : \frac{2}{9}

2 \cdot \frac{1}{9}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{9}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{9}

= \frac{\frac{4}{3} + \frac{2}{3}}{\frac{2}{9}}

Join

\frac{4}{3} + \frac{2}{3} : 2

\frac{4}{3} + \frac{2}{3}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{4 + 2}{3}

Add the numbers:

4 + 2 = 6

= \frac{6}{3}

Divide the numbers:

\frac{6}{3} = 2

= 2

= \frac{2}{\frac{2}{9}}

Apply the fraction rule:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{2 \cdot 9}{2}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{18}{2}

Divide the numbers:

\frac{18}{2} = 9

= 9

v = \frac{- ( - \frac{4}{3} ) - \frac{2}{3}}{2 \frac{1}{9}} : 3

\frac{- ( - \frac{4}{3} ) - \frac{2}{3}}{2 \cdot \frac{1}{9}}

Apply rule

- (- a) = a

= \frac{\frac{4}{3} - \frac{2}{3}}{2 \cdot \frac{1}{9}}

Multiply

2 \cdot \frac{1}{9} : \frac{2}{9}

2 \cdot \frac{1}{9}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{9}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{9}

= \frac{\frac{4}{3} - \frac{2}{3}}{\frac{2}{9}}

Join

\frac{4}{3} - \frac{2}{3} : \frac{2}{3}

\frac{4}{3} - \frac{2}{3}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{4 - 2}{3}

Subtract the numbers:

4 - 2 = 2

= \frac{2}{3}

= \frac{\frac{2}{3}}{\frac{2}{9}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{2 \cdot 9}{3 \cdot 2}

Cancel the common factor:

2

= \frac{9}{3}

Divide the numbers:

\frac{9}{3} = 3

= 3

The solutions to the quadratic equation are:

v = 9, v = 3

v = 9, v = 3

Substitute back

v = 3^{u},

solve for

u

Solve

3^{u} = 9 : u = 2

3^{u} = 9

Apply exponent rules

3^{u} = 9

Convert to base

3 : 3^{u} = 3^{2}

Convert

9

to base

3

9 = 3^{2}

3^{u} = 3^{2}

3^{u} = 3^{2}

a^{f (x)} = a^{g (x)}

, then

f (x) = g (x)

u = 2

u = 2

Solve

3^{u} = 3 : u = 1

3^{u} = 3

Apply exponent rules

3^{u} = 3

a^{f (x)} = a^{g (x)}

, then

f (x) = g (x)

u = 1

u = 1

u = 2, u = 1

Substitute back

u = sin (x)

sin (x) = 2, sin (x) = 1

sin (x) = 2, sin (x) = 1

sin (x) = 2 :

No Solution

sin (x) = 2

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn

Popular Examples

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5 sin (3 x) = - 2

cos^2(x)-1=0,0<= x<2pi

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Frequently Asked Questions (FAQ)

What is the general solution for 9^{sin(x)-1}+3=3^{sin(x)}+3^{sin(x)-1} ?
The general solution for 9^{sin(x)-1}+3=3^{sin(x)}+3^{sin(x)-1} is x= pi/2+2pin