Find values of y such that: log2(11y–3)–log2(3) –2log2(y) = 1

NB.: Treat all log as log2 for purpose of formatting log(x) - log(z) = log(x/z) alog(b) = log(b^a) log((11y - 3)/3) - log(y^2) = 1 log((11y - 3)/3y^2) = 1 11y - 3 / 3y^2 = 2^1 11y - 3 = 6y^2 6y^2 - 11y + 3 = 0 (3y - 1) (2y - 3) = 0 y = 1/3 or 1.5

SA
Answered by Shrinivas A. Maths tutor

5097 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do I find the equation of the normal line given a point on the curve?


How to differentiate e^x . sin(x)


How do I find the equation of a tangent to a given point on a curve?


Curve C has equation y=(9+11x)/(3-x-2x^2). Find the area of the curve between the interval (0, 1/2). State your answer in exact terms.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences