Ellie B. GCSE Maths tutor, 11 Plus Maths tutor, A Level Maths tutor, ...

Ellie B.

Currently unavailable: for regular students

Degree: Mathematics (Masters) - York University

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About me

Hi, my name is Ellie and I am currently studying Mathematics at the University of York. I chose York because I felt a real connection with the teachers, they made maths fun, interesting and easy. That's what I hope to do with all my students, maths at first can seem really quite daunting, but sometimes all you need is someone to just explain a problem in a different way and it all suddenly clicks into place. I know what it is like not to understand something and it can be really quite stressful, so I hope I can help students feel relaxed and confident in their subjects. 

My approach to teaching is that practice makes perfect, so I'll give my students many questions, this way I know what exactly they're really struggling with and then I can go back and teach them the fundamentals of that subject area. I also believe in giving my student's questions with a wide range of difficulties-as that is exactly what the exam does-and after-all to understand the harder questions you must understand the easiest ones. 

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Further Mathematics GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr
Maths 11 Plus £18 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA
Further MathematicsA-LevelA
PhysicsA-LevelB
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for regular students

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Questions Ellie has answered

Solve Inx + In3 = In6

To solve Inx + In3 = In6 we must follow some basic log rules,  logb(mn) = logb(m) + logb(n) if we compare this with the left side of our equation, Inx + In3, we will set m = x and n = 3,  mn is therefore 3x this means that Inx + In3 is equivalent to In3x So replacing that into our original ...

To solve Inx + In3 = In6 we must follow some basic log rules, 

logb(mn) = logb(m) + logb(n)

if we compare this with the left side of our equation, Inx + In3, we will set m = x and n = 3, 

mn is therefore 3x

this means that Inx + In3 is equivalent to In3x

So replacing that into our original equation:

In3x = In6

Take In of both sides

3x = 6

therefore x = 2

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4 months ago

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