(i messed up) solve the exponential equation algebraically.?
5e^(-0.04x) + 67 = 97 5e^(-0.04x) = 30 e^(-0.04x) = 6 ln e^(-0.04x) = ln 6 -0.04x = ln 6 x = ln 6 / -0.04 7^x = 5^(x - 4) ln 7^x = ln 5^(x - 4) x * ln 7 = (x - 4) * ln 5 x * ln 7 = (x * ln 5) - (4 * ln 5) (x * ln 5) - (x * ln 7) = (4 * ln 5) x(ln 5 - ln 7) = (4 * ln 5) x = (4 * ln 5) / ln(5/7)
How would you solve this equation algebraically: (7/x-4)=1+(9/x+4)?
(7/x-4) = 1 + (9/x+4) Do you mean 7/(x-4) = 1 + 9/(x+4)? If so, 7/(x-4) = 1 + 9/(x+4) 7/(x-4) = (x+4)/(x+4) + 9/(x+4) 7/(x-4) = (x+4+9)/(x+4) 7/(x-4) = (x+13)/(x+4) (x-4)(x+13) = 7(x+4) (multiplied both sides by (x-4) and (x+4)) x^2 + 13x - 4x - 52 = 7x + 28 x^2 + 9x - 52 = 7x + 28 x^2 + 9x - 7x - 52 - 28 = 0 x^2 - 2x - 80 = 0 (x - 10)(x + 8) = 0 x = 10 or -8
How to solve exponential equations algebraically? 7^x=5^(x-4)?
7^x = 5^(x - 4) Take the natural log of both sides: ln(7^x) = ln(5^(x - 4)) Use the log property that says log(a^b) = b • log(a): x • ln(7) = (x - 4) • ln(5) Distribute the ln(5): x • ln(7) = xln(5) - 4ln(5) Subtract xln(5) from goth sides: xln(7) - xln(5) = -4ln(5) Factor out the x: x[ln(7) - ln(5)] = -4ln(5) Divide both sides by ln(7) - ln(5): x = -4ln(5) / [ln(7) - ln(5)] x ≈ -19.13308
Exponential Equation help. Solve the equation for x: e^4x-e^2x=-6?
My teacher taught us to substitute e^2x as w since the equation is of quadratic type. So of course you move the -6 to the other side, giving you w^2-w+6. However, this cannot be factored and when I used the quadratic formula the answers I came up with were 1+5i/2 or 1-5i/2. The next step would be to plug that answer into the original equation for e^2x, but I have no idea how to do that. Did I do something wrong with what I did or is this problem really that tricky? Please either tell me my mistake or show me how you would plug the two answers I got back into the equation to solve for x. Thanks!
How do I solve this exponential equation 2^2x *2 =3^x *3^5?The left hand side is [math]2^{2x+1}[/math] and the right hand side is [math]3^{x+5}[/math].If you take logs, to any base you get [math](2x+1)\log(2) = (x+5)\log(3)[/math]. This is a linear equation that you shouldn’t need any further help with.
Algebraically solve the equation ln(x+2) -1 = ln(x-2).?
ln(x + 2) - 1 = ln(x - 2) ln(x + 2) - ln(e) = ln(x - 2) ln(x + 2) = ln(x - 2) + ln(e) ln(x + 2) = ln e(x - 2) Therefore, x + 2 = e(x - 2) x + 2 = ex - 2e 2 + 2e = ex - x 2 + 2e = x(e - 1) => x = (2 + 2e) / (e - 1)
[math]\log_2 (x+4)=3-x[/math][math]x+4=2^{3-x}[/math]Dividing through by [math]2^{3-x}[/math],[math](x+4)2^{x-3}=1[/math]Multiplying by [math]2^7=128[/math],[math](x+4)2^{x+4}=128[/math]We can convert the power into [math]e^y[/math] form:[math](x+4)e^{(x+4)\ln(2)}=128[/math]Multiplying through by [math]\ln(2)[/math],[math](x+4)\ln(2)\cdot e^{(x+4)\ln(2)}=128\ln(2)[/math]Now, we can apply the Lambert W Function to both sides:[math](x+4)\ln(2)=W(128\ln(2))[/math][math]x+4=\dfrac{W(128\ln(2))}{\ln(2)}[/math][math]x=\dfrac{W(128\ln(2))}{\ln(2)}-4[/math]The W function is a multivalued, transcendental, complex function. There are infinitely many complex solutions for [math]x[/math], and none of them have any nice closed form representation. It has a single real value, at [math]x=0.75158972705493911012[/math].
Try to relate the two terms with one another.What manipulations could you do to either one of them to obtain the other?So, the standard trick would be, to realise that [math](2 +\sqrt{3})[/math] and [math](2-\sqrt {3})[/math] conjugates.Which is to say that their product is rational! (and coincidentally 1)Let [math]2+\sqrt{3}[/math] be [math]p[/math].[math]\implies 2-\sqrt{3}= \frac {1}{p}[/math]Now the exponential equation can be reduced to:[math]\implies p^{x} + \frac {1}{p^{x}} = 4[/math][math]\implies {(p^{x})}^2 + 1 = 4.p^{x}[/math]This is a quadratic (YAY!) in [math]p^{x}[/math]- much easier to solve.Whatever roots you get equate to [math]{(2 + \sqrt{3})}^{x}[/math] - but remember that you can not have a negative solution to the quadratic!(Answer: 1,-1)
Can anyone solve the exponential equation 5^x+2=4^1-x?
one solution is (assuming you meant 5^(x+2) = 4^(1-x)) x = -Log(25/4) / (Log(4)+Log(5)) whether that's all solutions or not, i'm not sure... how would you solve it to begin with? well, 5^(x+2) = 4^(1-x) 25*5^x = 4*(1/4)^x 25/4 = (1/4)^x*(1/5)^x = (1/20)^x take the natural log of both sides... Log(25/4) = x*(Log(1/20)) = x*(-Log(20)) or x = -Log(25/4) / Log(20) or x = -Log(25/4) / (Log(4) + Log(5)) or of course x = Log(4/25) / (Log(4) + Log(5))