Find x
1) 6log(down)10 x=1.6
2) e^-4x=1.943
3) 1.3x10^-4 x=4.7/x -6.3x10^-3
4) 6lnx=1.6
5) x^2¡Ñ23y^2 = 1.78x10^32¡Ñz
please help I really can't solve this problem.
6log_10 x=1.6
log_10 x = 1.6/6
x = 10^(1.6/6) = 1.848
e^(-4x) = 1.943
-4x = ln 1.943
x = (ln 1.943)/-4 = -0.166
OK. I'll take a stab at parsing this one:
(1.3*10^-4) x = 4.7/(x - 6.3*10^-3)
x(x - 6.3*10^-3) = 4.7/(1.3*10^-4)
x^2 - 0.0063x - 36153.846 = 0
x = -190.138, 190.145
I suspect I'm wrong, but all those x's and missing parentheses gave me the willies...
6lnx = 1.6
This is just #1, but using base e instead of 10
Can't parse the special symbols. Try using some ascii characters, or even (ugh) words
I'd be happy to help you with these problems!
1) To find x in the equation 6log(down)10 x = 1.6, you can start by rewriting the equation in exponential form.
The logarithm equation log(down)10 x = y is equivalent to 10^y = x.
So, in this case, we have 10^1.6 = x. Evaluating this expression gives x ≈ 39.81.
2) To solve the equation e^-4x = 1.943 for x, you can take the natural logarithm (ln) of both sides of the equation.
This gives ln(e^-4x) = ln(1.943). Using the property of logarithms that ln(e^y) = y, we get -4x = ln(1.943).
To find x, divide both sides of the equation by -4: x = ln(1.943) / (-4). Evaluating this expression gives x ≈ -0.043.
3) Re-arranging the equation 1.3x10^-4 x = 4.7/x - 6.3x10^-3, we get the equation 1.3x10^-4 x^2 = 4.7 - 6.3x10^-3x.
Since this is a quadratic equation, we can set it equal to zero and solve it using the quadratic formula.
The quadratic formula is x = (-b ± √(b^2 - 4ac)) / (2a), where the equation is ax^2 + bx + c = 0.
In this case, a = 1.3x10^-4, b = 6.3x10^-3, and c = -4.7. Plugging these values into the quadratic formula will give you the solutions for x.
4) Similar to problem (2), to solve the equation 6lnx = 1.6 for x, you can divide both sides of the equation by 6: lnx = 1.6/6.
Taking the exponential of both sides gives e^(lnx) = e^(1.6/6), which simplifies to x = e^(1.6/6). Evaluating this expression gives x ≈ 1.246.
5) The equation x^2√23y^2 = 1.78x10^32√z can be a little trickier to solve. Since it involves multiple variables, we can't isolate x or y directly.
To solve for one variable, we need additional equations or information for the other variables. If you have any other equations or constraints, please let me know so that I can assist you further.
I hope this explanation helps you solve these problems! Let me know if you have any further questions.