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Solve the equation:
Solution
Note that 27, 9 and 3 may be written as powers of 3 as follows:
27 = 3^{3} , 9 = 3^{2} and 3 = 3^{1}
Using the above and also the formula \( \dfrac{1}{x^n} = x^{n} \), we rewrite the given equation as follows:
(3^{3})^{2x} (3^{2})^{x  2} = (3^{2})^{x} (3^{1})^{2  x}
We now use the formula (x^{m})^{n} = x ^{m n} to rewrite the above equation as follows.
3^{6x} 3^{2x + 4} = 3^{2x} 3^{ 2 + x}
Use the formula x^{m} x^{n } = x^{m + n} to rewrite the equation as follows
3^{6x 2x + 4} = 3^{ 2x  2 + x}
Simplify the exponents to obtain
3^{4x + 4} = 3^{ x  2}
We use the fact that an exponential function of the form a^{x} is a one to one function, meaning that if 3^{m} = 3^{n}, then m = n to write an algebraic equation using the above equation:
4x + 4 =  x  2
Solve the above for x to obtain the solution of the given equation.
5x =  6 gives the solution x =  6 / 5 =  1.2
Below is shown the graph of the left side of the given equation when written with right side equal to zero as follows: \( 27^{2x} \left( \dfrac{1}{9} \right)^{x2}  9^{x} \left( \dfrac{1}{3} \right)^{2x} = 0 \). The x intercept is an approximation to the analytical solution found above. Check that they are close in value.
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Find the solutions to the equation $$ \sqrt{16^x} = \left( \dfrac{1}{2} \right)^{x^21}. $$
Solution
Note that √(16^{x}) = (16^{x})^{1/2} = 16^{(1/2)x} and use it to rewrite the given equation as
16^{(1/2)x} = (1/2)^{x21}
Note that 1/2 = 2^{1} and 16 = 2^{4} and use it to rewrite the above equation as
(2^{4})^{(1/2)x} = (2^{1})^{x21}
Use the formula (x^{m})^{n} = x ^{m n} to rewrite the above equation as
2^{2x} = 2^{x2 + 1}
We use the fact that an exponential function of the form a^{x} is a one to one function to write.
2x =  x^{2} + 1
Rewrite in standard form and solve the above quadratic equation.
x^{2} + 2x  1 = 0
Δ = 2^{2}  4 (1)(1) = 8
Two solutions: x_{1} =  1  √ 2 ≈ 2.41 and x_{2} =  1 + √ 2 ≈ 0.41
The graph of the left side of the given equation when written with right side equal to zero as follows: \( \sqrt{16^x}  \left( \dfrac{1}{2} \right)^{x^21} = 0 \) is shown below. The x intercept are approximations to the solutions analytical solutions found above. Check that they have close values.
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What are the solutions to the equation $$ 2^x  6 \cdot 2^{x} = 6 ? $$
Solution
Let u = 2^{x} which also gives 1/u = 2^{ x} and rewrite the equation in terms of u.
u  6/u = 6
u cannot be zero (because 2^{x} cannot be less than or equal to zero), we therefore multiply all terms of the above equation by u and simplify
u(u  6/u) = 6 u
u^{2}  6 = 6 u
u^{2}  6u  6 = 0
Solve for u
Δ = (6)^{2}  4(1)(6) = 60
Two solutions: u_{1} = 3 + √15 and u_{2} = 3  √15
We now solve for x using the above substitution u = 2^{x}.
First equation: 2^{x} = 3 + √15
ln(2^{x}) = ln(3 + √15)
x ln(2) = ln(3 + √15)
Solution: x = ln(3 + √15) / ln 2 ≈ 2.78
Second equation: 2^{x} = 3  √15 has no solution because 3  √15 is less than zero.
Below is shown the graph of the left side of the given equation when written with right side equal to zero as follows: \( 2^x  6 \cdot 2^{x}  6 = 0 \). The x intercept is an approximation to the solution of the analytical solution found above.
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Find the solutions to the equation $$ 5^{x+1} = 100 \cdot 3^x . $$
Solution
Take the natural logarithm (ln) of both sides
ln (5^{x + 1}) = ln (100 ⋅ 3^{x} )
Use log rules to simplify.
(x + 1) ln 5 = ln 100 + x ln 3
Expand left side and group terms with x.
x ln 5 + ln 5 = ln 100 + x ln 3
x (ln 5  ln 3) = ln 100  ln 5
x = (ln 100  ln 5) / (ln 5  ln 3) ≈ 5.86
Below is shown the graph of the left side of the given equation when written with right side equal to zero as follows: \( 5^{x+1}  100 \cdot 3^x = 0 \). The x intercept is an approximation to the solution of the analytical solution found above.
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Solve the equation: $$ \dfrac{e^xe^{x}}{2} = 3 $$
Solution
Let u = e^{x} which also gives 1/u = e^{ x} and rewrite the equation in u. (u = e^{x} cannot be less than or equal to zero)
(u  1/u) / 2 = 3
Multiply both sides by 2 and simplify
u  1 / u = 6
Multiply both sides by u and simplify
u^{2}  1 = 6 u
u^{2}  6u  1 = 0
Solve for u
Δ = (6)^{2}  4(1)(1) = 40
Two solutions: u_{1} = 3 + √10 and u_{2} = 3  √10
We now solve for x using the above substitution u = e^{x}.
First equation: e^{x} = 3 + √10
ln(e^{x}) = ln(3 + √10)
Solution: x = ln(3 + √10) ≈ 1.82
Second equation: e^{x} = 3  √10 has no solution because 3  √10 is less than zero.
Below is shown the graph of the left side of the given equation when written with right side equal to zero as follows: \( \dfrac{e^xe^{x}}{2}  3 = 0 \). The x intercept is an approximation to the solution of the analytical solution found above.
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What are the solutions to the equation $$ 3^{23x} = 4^{2x+1} \,\,?$$
Solution
Take the natural logarithm (ln) of both sides
ln (3^{2  3x}) = ln (4^{2x + 1})
Use log rules to simplify.
(2  3x) ln 3 = (2x + 1) ln 4
Expand both sides of the equation and group terms with x.
2 ln 3  3 x ln 3 = 2 x ln 4 + ln 4
x (3 ln 3  2 ln 4) = ln 4  2 ln 3
x = (2 ln 3  ln 4) / (3 ln 3 + 2 ln 4) ≈ 0.13
Below is shown the graph of the left side of the given equation when written with right side equal to zero as follows: \( 3^{23x}  4^{2x+1} = 0 \). The x intercept is an approximation to the solution of the analytical solution found above.
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Solve for x the equation: $$ 9^x −3^{x+1} + 2 = 0 $$
Solution
Note that
9^{x} = (3^{2})^{x} = 3^{2 x} = (3^{x})^{2}
Rewrite the given equation using 9^{x} = (3^{x})^{2} and also 3^{x + 1} = 3⋅3^{x}
(3^{x})^{2}  3⋅3^{x} + 2 = 0
Let u = 3^{x} and rewrite the equation in u.
u^{2}  3 u + 2 = 0
Solve for u by factoring
(u  2)(u  1) = 0
Two solutions: u_{1} = 2 and u_{2} = 1
We now solve for x using the above substitution u = 3^{x}.
First equation: 3^{x} = 2
ln( 3^{x}) = ln 2
x ln 3 = ln 2
Solution: x = ln 2 / ln 3 ≈ 0.63
Second equation: 3^{x} = 1 , solution x = 0.
Below is shown the graph of the left side of the given equation. The x intercepts are approximations to the analytical solutions found above.
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