A few logarithm questions (1 Viewer)

[minijumbuk]

My first time posting question for maths xD

1. If (2^m)ⁿ = (8ⁿ)^m, and m,n≥0 and n=/=1, find m in terms of n.

Answer: (3n)^1/(n-1)

2. 2^x+y = 6^y , 3^x = 6(2^y)
Answer to 4 d.p.

Answer: x= 2.7097, y = 1.7096

3. Express in rational denominator:
(√5 - √3) / (√2 + √3 + √5 )

Answer: (√15 + 3√10 - 4√6 - 3) / 6

4. If (log₅a)(log_a2a)(log_2a4a)(log_4ab)=2 and a>1, b>1, find the value of b.

Answer: b=25

As you can see, I've got the answers. I just don't know how to get to them =\

Thanks in advance!

 

[3unitz]

1)
is it suposed to be 2^(m^n) = (8^n)^m?
2^(m^n) = (2^3n)^m
2^[m^(n-1)] = 2^3n
m^(n-1) = 3n
m = (3n)^[1/(n-1)]

2)
2^(x + y) = 6^y
(x+y)ln2 = yln6
xln2 + yln2 = yln6
x = y(ln6 - ln2)/ln2
x= y ln3/ln2 -----------(1)

3^x = 6(2^y)
xln3 = ln6 + yln2
sub in (1):
(y ln3/ln2)ln3 = ln6 + yln2
y (ln3)^2/ln2 - yln2 = ln6
y[(ln3)^2/ln2 - ln2] = ln6
y = ln6 / [(ln3)^2/ln2 - ln2]
y = 1.7096

sub y into (1):
x = 2.7097

3)
consider multiplying the denominator by [(r2 + r3) - r5]:
[(r2 + r3) + r5] [(r2 + r3) - r5]
= (r2 + r3)^2 - (r5)^2
= 2 + 2.r2.r3 + 3 - 5
= 2 r6

consider multiplying 2 r6 by (r6/2):
2r6 (r6/2) = 6 <---- denominator now rationalised

hence now multiply the numerator by [(r2 + r3) - r5](r6/2):
(r5 - r3)[(r2 + r3) - r5](r6/2)
= (r5 - r3)[r2 - (r5 - r3)](r6/2)
= [r2(r5 - r3) - (r5 - r3)^2](r6/2)
= (r10 - r6 - 5 + 2r15 - 3)(r6/2)
= r60/2 - 6/2 - 5r6/2 + 2r90/2 - 3r6/2
= r60/2 + r90 - 4r6 - 3
= r15 + 3r10 - 4r6 - 3

.'. answer: (r15 + 3r10 - 4r6 - 3) / 6

4)
(ln a / ln 5)(ln 2a / ln a) (ln4a / ln2a)(ln b/ln4a) = 2 (by change of base)
(ln a / ln 5)(ln 2a / ln a) (ln4a / ln2a)(ln b/ln4a) = 2
ln b / ln 5 = 2
ln b = 2 ln5
ln b = ln 25
b = 25

 

[lyounamu]

My first time posting question for maths xD

1. If (2^m)ⁿ = (8ⁿ)^m, and m,n≥0 and n=/=1, find m in terms of n.

Answer: (3n)^1/(n-1)

2. 2^x+y = 6^y , 3^x = 6(2^y)
Answer to 4 d.p.

Answer: x= 2.7097, y = 1.7096

3. Express in rational denominator:
(√5 - √3) / (√2 + √3 + √5 )

Answer: (√15 + 3√10 - 4√6 - 3) / 6

4. If (log₅a)(log_a2a)(log_2a4a)(log_4ab)=2 and a>1, b>1, find the value of b.

Answer: b=25

As you can see, I've got the answers. I just don't know how to get to them =\

Thanks in advance!

1) Your solution does not work.

 

[lyounamu]

1)
is it suposed to be 2^(m^n) = (8^n)^m?
2^(m^n) = (2^3n)^m
2^[m^(n-1)] = 2^3n
m^(n-1) = 3n
m = (3n)^[1/(n-1)]

2)
2^(x + y) = 6^y
(x+y)ln2 = yln6
xln2 + yln2 = yln6
x = y(ln6 - ln2)/ln2
x= y ln3/ln2 -----------(1)

3^x = 6(2^y)
xln3 = ln6 + yln2
sub in (1):
(y ln3/ln2)ln3 = ln6 + yln2
y (ln3)^2/ln2 - yln2 = ln6
y[(ln3)^2/ln2 - ln2] = ln6
y = ln6 / [(ln3)^2/ln2 - ln2]
y = 1.7096

sub y into (1):
x = 2.7097

3)
consider multiplying the denominator by [(r2 + r3) - r5]:
[(r2 + r3) + r5] [(r2 + r3) - r5]
= (r2 + r3)^2 - (r5)^2
= 2 + 2.r2.r3 + 3 - 5
= 2 r6

consider multiplying 2 r6 by (r6/2):
2r6 (r6/2) = 6 <---- denominator now rationalised

hence now multiply the numerator by [(r2 + r3) - r5](r6/2):
(r5 - r3)[(r2 + r3) - r5](r6/2)
= (r5 - r3)[r2 - (r5 - r3)](r6/2)
= [r2(r5 - r3) - (r5 - r3)^2](r6/2)
= (r10 - r6 - 5 + 2r15 - 3)(r6/2)
= r60/2 - 6/2 - 5r6/2 + 2r90/2 - 3r6/2
= r60/2 + r90 - 4r6 - 3
= r15 + 3r10 - 4r6 - 3

.'. answer: (r15 + 3r10 - 4r6 - 3) / 6

4)
(ln a / ln 5)(ln 2a / ln a) (ln4a / ln2a)(ln b/ln4a) = 2 (by change of base)
(ln a / ln 5)(ln 2a / ln a) (ln4a / ln2a)(ln b/ln4a) = 2
ln b / ln 5 = 2
ln b = 2 ln5
ln b = ln 25
b = 25

How did you move from there to that line?

Great work by the way, you beat me to it. I was going to do Q3 and 4 as I could not do Q2 and the solution for Q1 didn't work.

 

[3unitz]

How did you move from there to that line?

took log of both sides:

2^(x + y) = 6^y

ln [2^(x + y)] = ln (6^y)

now used ln (a^b) = b ln a:

(x+y) ln2 = y ln6

 

[lyounamu]

took log of both sides:

2^(x + y) = 6^y

ln [2^(x + y)] = ln (6^y)

now used ln (a^b) = b ln a:

(x+y) ln2 = y ln6

Thanks. I got confused with ln with log10 for a while due to excessive amount of exercises I did on changing bases.

 

[minijumbuk]

I'm not sure, all these answers were given to me. They might've been wrong.

 

[minijumbuk]

Nice, thanks =)

Man I'm so hopeless at this =O
One of these questions would take me quite a while to grip the pattern.....=\

 

[lyounamu]

Nice, thanks =)

Man I'm so hopeless at this =O
One of these questions would take me quite a while to grip the pattern.....=\

Yeah, I reckon they were of quite high difficulty. I could only get Q3 and Q4 when I did it. I had no clue on Q2, lol.

 

[minijumbuk]

Hahaha.

I'm sure if we spent some more time we'd eventually get it... But we don't get to do that in the exam T_T

 

[lyounamu]

Hahaha.

I'm sure if we spent some more time we'd eventually get it... But we don't get to do that in the exam T_T

Yeah, I would hate to get those questions in exam. That would be extremely time-consuming.