Solve the following exponential equation: [math]2^ {2x+1} = 2^{9}[/math]?
I put the math parts of the question into math mode. Let’s see if that’s enough editing to be improved.This one is easy.[math]2^{2x+1}=2^9[/math]Taking logs base two just means we can equate the exponents:[math]2x+1=9[/math][math]2x=8[/math][math]x=4[/math]Check: [math]2^{2(4)+1}=2^9 \ \ \checkmark[/math]
How do I solve this exponential equation [math](2+\sqrt 3) ^x + (2-\sqrt 3) ^x = 4[/math]?
[math](2+ \sqrt{3})^x + (2 - \sqrt{3})^x = 4[/math][math](2 + \sqrt{3})^{2x} + (4 - 3)^x = 4(2 + \sqrt{3})^x[/math][math](2 + \sqrt{3})^{2x} + 1 = 4(2 + \sqrt{3})^x[/math][math](2 + \sqrt{3})^{2x} - 4(2 + \sqrt{3})^x + 1 = 0[/math][math]((2 + \sqrt{3})^{2x} - 4(2 + \sqrt{3})^x + 4) - 3 = 0[/math][math]((2 + \sqrt{3})^x - 2)^2 = 3[/math][math](2 + \sqrt{3})^x - 2 = \pm \sqrt{3}[/math][math](2 + \sqrt{3})^x = 2 \pm \sqrt{3}[/math][math](2 + \sqrt{3})^x = 2 + \sqrt{3}[/math][math](2 + \sqrt{3})^x = (2 + \sqrt{3})^1[/math][math]x = 1[/math][math](2 + \sqrt{3})^x = 2 - \sqrt{3}[/math][math](2 + \sqrt{3})^{x + 1} = 4 - 3[/math][math](2 + \sqrt{3})^{x + 1} = 1[/math][math](2 + \sqrt{3})^{x + 1} = (2 + \sqrt{3})^0[/math][math]x + 1 = 0[/math][math]x = -1[/math][math]x = -1, 1[/math]
How do you solve an exponential equation given e^x=4?
To solve the given exponential equation for x, e^x = 4, you take the natural logarithm of both sides of the equation as follows:(1.) ln e^x = ln 4Remember: Logarithms are exponents, and in the case of natural logarithms (ln), i.e., logarithms to the base e, y = ln x says, "To what power or exponent must I raise the irrational number e = 2.718 (rounded to 3 decimal places) in order to get or produce x?" Answer: y; therefore, when I want to find the natural logarithm of e^x, I ask myself the same question, "To what power or exponent must I raise the irrational number e in order to get e^x?" Answer: it is obviously x, that is, ln e^x = x; Remember: Logarithms are exponents! Therefore, equation (1.) becomes:(2) x = ln 4 = 1.38629436 (rounded to 7 decimal places)CHECK:e^x = 4e^(ln 4) = 4e^(1.38629436) is approximately equal to (≈) 4 due to rounding.3.9999999955 ≈ 4Therefore, x = ln 4 is indeed the exact answer or solution to the given exponential equation.
Solve for x in each of the following exponential equations:?
1) 7^2 = 49 So 2x= 2 x=1 2) 2^3x = 1/64 2^ 1/6 = 1/64 so 3x= 1/6 x= 1/18 3) 6^-x = 1/216 6 ^ -1/3 = 1/216 So -x = -1/3 x= 1/3 4) 3^(x-1) = 81 3^4 = 81 So x-1 = 4 x=5 5) 9^2x = 27 3^4x= 27 3^3 = 27 So 4x = 3 x = 3/4 6) 5^(5x-2) = 125 5^2 = 125 So 2 = 5x-2 5x= 4 x= 4/5 7) 36^(x-3) = 216 6^2(x-3) = 216 6^3 = 216 So 3= 2x-6 2x=9 x=9/2 8) 3^x = 27^(2 - 3x) 3^x = 3^3(2-3x) x= 6-9x 10x= 6 x = 3/5 9) 4^x = 1/16 4^ 1/2 = 1/16 1/2 = x 10) 27^(1-x) =3 27^ 1/3 = 3 1-x = 3 x = -2 11) What is eleven supposed to equal? Is it 4^(x+1) = 2^x 2^ 2(x+1) = 2^x 2x+2 = x x=-2 12) 9^(1-x) = 27^(x-1) 3^2(1-x) = 3^3(x-1) 2-2x = 3x-3 5x=5 x=1 Hope I got all of those right. That was a long one
I want to solve these exponential equations. Any help?
1. 19^x = 143 x = (ln 143) / (ln 19) x =~ 1.69 2. 2^(2x) + 2^x - 12 = 0 (2^x)^2 + 2^x - 12 = 0 Let u = 2^x, so we have: u^2 + u - 12 = 0 u^2 + 4u - 3u - 12 = 0 u(u + 4) - 3(u + 4) = 0 (u - 3)(u + 4) = 0 u - 3 = 0 or u + 4 = 0 u = 3 or u = -4 2^x = 3 or 2^x = -4 The latter case is impossible, so we have: 2^x = 3 x = (ln 3) / (ln 2) x =~ 1.58 3. x - 7 = 5^2 x - 7 = 25 x = 25 + 7 x = 32 4. 3 log x = log 125 log 125 / log x = 3 x^3 = 125 x = 5 5. log (x - 2) + log 5 = log 100 (x - 2)(5) = 100 5x - 10 = 100 5x = 100 + 10 5x = 110 x = 110/5 x = 22
How do you solve an exponential equation with a variable in the base?
How do you solve an exponential equation with a variable in the base?You mean something like [math]x^a=b[/math] with [math]a[/math] and [math]b[/math] given numbers? Simply take the [math]a[/math]-root from [math]b[/math]. [math]x=\sqrt[a]{b}=b^{\frac{1}{a}}[/math].More general “exponential equation with a variable in the base” are called polynomials and you can solve them up to the 4th degree (4 is the highest exponent and only natural exponents), now solving a polynomaial of the first degree (linear equation) and second degree (quadric equation) is easy, for 3 and 4 degree you can surely find methods by looking.
2^x+8=5 Solve the exponential equation. Express the solution set in terms of natural logarithms?
If 2^(x+8)=5 then x+8 = log(5)/log(2) then x = log(5)/log(2) - 8 If (2^x) + 8 = 5 then no solution because 2^x = -3 never (x real number) Thanks for low rating Result is OK = log(5)/log(2) - 8 = -5.678071905 the same as = (LN(5) - 8*LN(2))/LN(2) = -5.678071905
2 Solving Exponential Equations? please help?
(√7)^(6x) = 49^(x - 2) [7^(1/2)]^(6x) = (7^2)^(x - 2) 7^(6x/2) = 7^2(x - 2) 7^(3x) = 7^(2x - 4) 3x = 2x - 4 x = -4 8 = 2^(x + 4) 2^3 = 2^(x + 4) 3 = x + 4 x = 3 - 4 x = -1 1/36 = 6^(x - 3) 1/(6^2) = 6^(x - 3) 6^(-2) = 6^(x - 3) -2 = x - 3 x = -2 + 3 x = 1