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(625^{cos^2(x)})/(25^{cos(x))}=1

Solution

\frac{62 5 ^{c o s^{2} (x)}}{2 5 ^{c o s (x)}} = 1

Solution

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

\frac{62 5 ^{c o s^{2} (x)}}{2 5 ^{c o s (x)}} = 1

Solve by substitution

\frac{62 5 ^{c o s^{2} (x)}}{2 5 ^{c o s (x)}} = 1

Let:

cos (x) = u

\frac{62 5 ^{u^{2}}}{2 5 ^{u}} = 1

\frac{62 5 ^{u^{2}}}{2 5 ^{u}} = 1 : u = 0, u = \frac{1}{2}

\frac{62 5 ^{u^{2}}}{2 5 ^{u}} = 1

Apply exponent rules

\frac{62 5 ^{u^{2}}}{2 5 ^{u}} = 1

Apply exponent rule:

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

\frac{1}{2 5 ^{u}} = 2 5^{- u}

62 5^{u^{2}} \cdot 2 5^{- u} = 1

Convert to base

25 : 2 5^{2 u^{2}} \cdot 2 5^{- u} = 1

Convert

625

to base

25

625 = 2 5^{2}

(2 5^{2})^{u^{2}} \cdot 2 5^{- u} = 1

Apply exponent rule:

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

(2 5^{2})^{u^{2}} = 2 5^{2 u^{2}}

2 5^{2 u^{2}} \cdot 2 5^{- u} = 1

2 5^{2 u^{2}} \cdot 2 5^{- u} = 1

Apply exponent rule:

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

2 5^{2 u^{2}} \cdot 2 5^{- u} = 2 5^{2 u^{2} - u}

2 5^{2 u^{2} - u} = 1

f (x) = g (x)

, then

ln (f (x)) = ln (g (x))

ln (2 5^{2 u^{2} - u}) = ln (1)

Apply log rule:

ln (x^{a}) = a \cdot ln (x)

ln (2 5^{2 u^{2} - u}) = (2 u^{2} - u) ln (25)

(2 u^{2} - u) ln (25) = ln (1)

(2 u^{2} - u) ln (25) = ln (1)

Solve

(2 u^{2} - u) ln (25) = ln (1) : u = 0, u = \frac{1}{2}

(2 u^{2} - u) ln (25) = ln (1)

Factor

(2 u^{2} - u) ln (25) : 2 ln (5) u (2 u - 1)

(2 u^{2} - u) ln (25)

Factor

2 u^{2} - u : u (2 u - 1)

2 u^{2} - u

Factor out common term

u : u (2 u - 1)

2 u^{2} - u

Apply exponent rule:

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

u^{2} = uu

= 2 uu - u

Factor out common term

u

= u (2 u - 1)

= u (2 u - 1)

= u (2 u - 1) ln (25)

Simplify

ln (25) : 2 ln (5)

ln (25)

Rewrite

25

in power-base form:

25 = 5^{2}

= ln (5^{2})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (5^{2}) = 2 ln (5)

= 2 ln (5)

= 2 ln (5) u (2 u - 1)

2 ln (5) u (2 u - 1) = ln (1)

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or 2 u - 1 = 0

Solve

2 u - 1 = 0 : u = \frac{1}{2}

2 u - 1 = 0

Move

1

to the right side

2 u - 1 = 0

Add

1

to both sides

2 u - 1 + 1 = 0 + 1

Simplify

2 u = 1

2 u = 1

Divide both sides by

2

2 u = 1

Divide both sides by

2

\frac{2 u}{2} = \frac{1}{2}

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

The solutions to the quadratic equation are:

u = 0, u = \frac{1}{2}

u = 0, u = \frac{1}{2}

Verify Solutions:

u = 0

True

, u = \frac{1}{2}

True

Check the solutions by plugging them into

\frac{62 5 ^{u^{2}}}{2 5 ^{u}} = 1

Remove the ones that don't agree with the equation.

Plug in

u = 0 :

True

\frac{62 5 ^{0^{2}}}{2 5 ^{0}} = 1

\frac{62 5 ^{0^{2}}}{2 5 ^{0}} = 1

\frac{62 5 ^{0^{2}}}{2 5 ^{0}}

Apply rule

0^{a} = 0

0^{2} = 0

= \frac{62 5 ^{0}}{2 5 ^{0}}

Apply rule

a^{0} = 1, a \neq = 0

2 5^{0} = 1

= \frac{62 5 ^{0}}{1}

Apply rule

a^{0} = 1, a \neq = 0

= \frac{1}{1}

Apply rule

\frac{a}{1} = a

= 1

1 = 1

True

Plug in

u = \frac{1}{2} :

True

\frac{62 5 ^{(\frac{1}{2})^{2}}}{2 5 ^{(\frac{1}{2})}} = 1

\frac{62 5 ^{(\frac{1}{2})^{2}}}{2 5 ^{(\frac{1}{2})}} = 1

\frac{62 5 ^{(\frac{1}{2})^{2}}}{2 5 ^{\frac{1}{2}}}

2 5^{\frac{1}{2}} = 5

2 5^{\frac{1}{2}}

Factor the number:

25 = 5^{2}

= (5^{2})^{\frac{1}{2}}

Apply exponent rule:

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

(5^{2})^{\frac{1}{2}} = 5^{2 \cdot \frac{1}{2}} = 5

= 5

= \frac{62 5 ^{(\frac{1}{2})^{2}}}{5}

62 5^{(\frac{1}{2})^{2}} = 62 5^{\frac{1}{2 ^{2}}}

62 5^{(\frac{1}{2})^{2}}

(\frac{1}{2})^{2} = \frac{1}{2 ^{2}}

(\frac{1}{2})^{2}

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= 62 5^{\frac{1}{2 ^{2}}}

= \frac{62 5 ^{\frac{1}{2 ^{2}}}}{5}

2^{2} = 4

= \frac{62 5 ^{\frac{1}{4}}}{5}

62 5^{\frac{1}{4}} = 5

62 5^{\frac{1}{4}}

Factor the number:

625 = 5^{4}

= (5^{4})^{\frac{1}{4}}

Apply exponent rule:

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

(5^{4})^{\frac{1}{4}} = 5^{4 \cdot \frac{1}{4}} = 5

= 5

= \frac{5}{5}

Apply rule

\frac{a}{a} = 1

= 1

1 = 1

True

The solutions are

u = 0, u = \frac{1}{2}

Substitute back

u = cos (x)

cos (x) = 0, cos (x) = \frac{1}{2}

cos (x) = 0, cos (x) = \frac{1}{2}

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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