# horrible log questionWatch

#1
here's a difficult log question im stuck on:
4^(x + 2) = 5^x
x + 2 log10 4 = x log10 5

then what...
0
12 years ago
#2
(Original post by alison_141288)
here's a difficult log question im stuck on:
4^(x + 2) = 5^x
x + 2 log10 4 = x log10 5

then what...
I'm stuck on them too! I can't get the right answer, I think my steps are fine though, could someone help me with 5^ (7x-1)= 12 and 7^(3- 2x) = 19. Thanx!
0
12 years ago
#3
(Original post by alison_141288)
here's a difficult log question im stuck on:
4^(x + 2) = 5^x
x + 2 log10 4 = x log10 5

then what...
it should be (x+2) lg 4 = x lg 5

x lg 4 + 2 lg 4 = x lg 5

lg 16 = x lg 5 - x lg 4

lg 16 = x (lg 5-lg 4)

x = lg 16/ ( lg 5 - lg 4 )

mrm.
0
12 years ago
#4
(Original post by alison_141288)
here's a difficult log question im stuck on:
4^(x + 2) = 5^x
x + 2 log10 4 = x log10 5

then what...
(x + 2) log10 4 = x log10 5
2 log10 4 + x log10 4 = x log10 5
2 log10 4 = -x (log10 5 - log10 4)

x= 2 times log10 4 / (log10 5 - log10 4) = Ans
0
12 years ago
#5
(Original post by alison_141288)
here's a difficult log question im stuck on:
4^(x + 2) = 5^x
x + 2 log10 4 = x log10 5

then what...
(x+2)k = xc

x(k-c) = -2k

x= -2k/(k-c)

your logs are just constants, and you have a calculator
0
12 years ago
#6
4^(x + 2) = 5^x

(x+2)log4 = xlog5
(x+2) = x(log5/log4)
2 = x[(log5/log4) - 1)]
x = 2/[(log5/log4) - 1)]
0
12 years ago
#7
(Original post by Shehna)
I'm stuck on them too! I can't get the right answer, I think my steps are fine though, could someone help me with 5^ (7x-1)= 12 and 7^(3- 2x) = 19. Thanx!

5 ^(7x-1) = 12

(7x-1) lg 5 = lg 12

7x lg 5 - 1 lg 5 = lg 12

7x lg 5 = lg 12 - lg 5

x = (lg 12 - lg 5) / (7 lg 5)
0
12 years ago
#8
hmmmmm, these boards are quite busy on Sunday nights... O_O
0
12 years ago
#9
Thank you all! Wish I had more reps to give out, but you are all on my list
0
#10
think ive just about tackled logs. off to bed now. gd luck to everyone tomorrow - we're gonna ace it!!
0
12 years ago
#11
5^(7x - 1) = 12
(7x - 1)log5 = log12
(7x - 1) = log12/log5
7x = (log12/log5) + 1
x = [(log12/log5) + 1]/7

7^(3 - 2x) = 19
(3 - 2x)log7 = log19
3 - 2x = log19/log7
2x = 3 - (log19/log7)
x = [3 - (log19/log7)]/2
0
12 years ago
#12
(Original post by SunGod87)
5^(7x - 1) = 12
(7x - 1)log5 = log12
(7x - 1) = log12/log5
7x = (log12/log5) + 1
x = [(log12/log5) + 1]/7

7^(3 - 2x) = 19
(3 - 2x)log7 = log19
3 - 2x = log19/log7
2x = 3 - (log19/log7)
x = [3 - (log19/log7)]/2
Thanx a million! I know this is reallly cheeky but could you possibly show me how to do 5^(x-1) = 4^(1-3x) , thanx again
0
12 years ago
#13
(Original post by Shehna)
Thanx a million! I know this is reallly cheeky but could you possibly show me how to do 5^(x-1) = 4^(1-3x) , thanx again

(x-1) lg 5 = ( 1-3x) lg 4

x lg 5 - lg 5 = lg 4 - 3x lg 4

x lg 5 + 3x lg 4 = lg 4 + lg 5

x ( lg 5 + 3 lg 4) = lg 4 + lg 5

x = (lg 4 + lg 5) / ( lg 5 + 3 lg 4)
0
12 years ago
#14
(Original post by Mrm.)
(x-1) lg 5 = ( 1-3x) lg 4

x lg 5 - lg 5 = lg 4 - 3x lg 4

x lg 5 + 3x lg 4 = lg 4 + lg 5

x ( lg 5 + 3 lg 4) = lg 4 + lg 5

x = (lg 4 + lg 5) / ( lg 5 + 3 lg 4)

which I would tidy up to

x = lg (20) / ( lg 5 + lg 64)

x= lg 20 / lg 320
0
12 years ago
#15
(Original post by Mrm.)
(x-1) lg 5 = ( 1-3x) lg 4

x lg 5 - lg 5 = lg 4 - 3x lg 4

x lg 5 + 3x lg 4 = lg 4 + lg 5

x ( lg 5 + 3 lg 4) = lg 4 + lg 5

x = (lg 4 + lg 5) / ( lg 5 + 3 lg 4)
Thank you!
0
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