This section tests for knowledge of algebraic symbols, elementary functionsand their inverses.Solve for x. ln(x)23 = ya) x = e√3yb) x =q 3ylnc) x = ln(y)23d) x = √e3y
Question
This section tests for knowledge of algebraic symbols, elementary functionsand their inverses.Solve for x. ln(x)23 = ya) x = e√3yb) x =q 3ylnc) x = ln(y)23d) x = √e3y
Solution
The equation given is ln(x^23) = y.
To solve for x, we need to get rid of the natural logarithm (ln). We can do this by using the property of logarithms that states if ln(a) = b, then a = e^b.
Applying this to our equation, we get x^23 = e^y.
To isolate x, we take the 23rd root of both sides, which gives us x = (e^y)^(1/23).
So, the correct answer is not listed among the options you provided. The correct answer should be x = (e^y)^(1/23).
Similar Questions
If arcsin ln ,x y= then dydx =(A) 21yx-(B) 21xyx-(C) 21yx+(D) arcsin xe(E) arcsin21xex+
y = √25 − x2, y = 3
y=−log3(2x)𝑦=−log3(2𝑥)Answer 1 Question 9y=log3(x+1)−4𝑦=log3(𝑥+1)−4Answer 2 Question 9y=2log3x𝑦=2log3𝑥Answer 3 Question 9y=12log3(x)𝑦=12log3(𝑥)Answer 4 Question 9
Multiply the radical expressions and simplify your answer.−36x6y3−−−−−√⋅210xy6−−−−−√−36𝑥6𝑦3⋅210𝑥𝑦6Select one:a. −12x3y415xy−−−−√−12𝑥3𝑦415𝑥𝑦b. −1215x7y9−−−−−−√−1215𝑥7𝑦9c. −24x3y415xy−−−−√−24𝑥3𝑦415𝑥𝑦d. −24xyx5y7−−−−√
Find the inverse of the function y = e5x.On the same diagram sketch the line y = x, and the graphs ofy = e5x and its inverse.(b) Let y = 240 e−7x and set Y = ln y.Find an expression for Y in terms of x and then draw the graph of Yagainst x
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.