help with logarithmics?

Question

log7 (x^2 + 36) = log7 100

and

log2 (4y-10) ≥ log2 (y-1)

left my book at home...thanks guys!

Answer 1

Cancel the logs from both sides

x^2+36=100
x^2=64
x=8,-8

4y-10>=y-1
3y>=9
y>=3

Answer 2

If the log are equal the numbers are equal

x^2+36= 100 x^2= 64 and x=+-8

2nd
As the base of log is2 >1 The same inequality between logs is mantained by numbers.

But first 4y-10 >0 so y> 5/2 and y-1>0 so y>1
so y >5/2

4y-10>= y-1 so 3y>= 9 and y >=3

Next time take your books

Answer 3

log7 (x^2 + 36) = log7 100 take anti log of both sides
x^2+36=100 subr=tract 100 from each side
x^2-64=0
(x+8)(x-8)=0
x+8=0
x=-8
x-8=0
x=8

x=8, -8

log2 (4y-10) ≥ log2 (y-1) take antilog of both sides
4y-10≥y-1 add 10 to each side
4y≥y+9 subtract t from each side
3y≥9 divide both sides by 3
y≥3