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Logarithms...

What is "lg"?

Need helping solving this:

Solve the equation lg(5x+8) + lg(2x+1) - 2lg(x+1)

Help much appreciated... :smile:

Reply 1

kikzen

14

lg is log to the base 10, or the natural log. it depends.
in this case it doesnt really matter.

remember the log laws :
lgx + lgy = lgxy
lgx - lgy = lg(x/y)
nlgx = lg(x^n)

Spoiler

Reply 2

massimo

i got lg5x+lg8+log2x+log1-log(x+1)²

log80x²÷logx²+2x+1=log(80+40x+80x²)

log(2x²+1+2)

so x=-0.25±√15÷16
log₁₀n=-0.25+√15÷16
log₁₀n=-0.25-√15÷16
and then just plug it in and get the answer but i need confirmation of this as im not sure its right !!

Reply 3

yazan_l

5

actually: log and lg are to the base 10, and ln is to the base e.
but mathematians use log instead of ln, but they use lg for log to the base 10.

Reply 4

super_baros

lgeerthan

What is "lg"?

Need helping solving this:

Solve the equation lg(5x+8) + lg(2x+1) - 2lg(x+1)

Help much appreciated... :smile:

ermmm.. unless my teacher taught me badly..... (which i think she did :biggrin:

)... is that not an expression?.... :confused:

What is it equal to?...

Reply 5

steve2005

17

lgeerthan

What is "lg"?

Need helping solving this:

Solve the equation lg(5x+8) + lg(2x+1) - 2lg(x+1)

Help much appreciated... :smile:

This isn't an equation
lg(5x+8) + lg(2x+1) - 2lg(x+1)

I presume you mean
lg(5x+8) + lg(2x+1) = 2lg(x+1)

but this would give a quadratic with negative solutions. One of which is impossible. The other value for x is -0.45.....

Can you please check you have typed the question correctly?

Reply 6

kikzen

14

he probably just meant 'expression' and that hewanted it simplified or something? well thats what i thought.
who knows?!

Reply 7

super_baros

massimo

i got lg5x+lg8+log2x+log1-log(x+1)²

log80x²÷logx²+2x+1=log(80+40x+80x²)

log(2x²+1+2)

so x=-0.25±√15÷16
log₁₀n=-0.25+√15÷16
log₁₀n=-0.25-√15÷16
and then just plug it in and get the answer but i need confirmation of this as im not sure its right !!

well massimo does..... :p:

...(no offence!)...

Reply 8

steve2005

17

super_baros

well massimo does..... :p:

...(no offence!)...

The value of x is -0.47.... I have plotted the LHS =y and the RHS =y the graphs intersect. I have also plotted the lg(5x............ =y this graph crosses the x-axis at -0.47.....

Solve the quadratic ( pic 14)

well i answered the question as written and i think its the right answer so far unless this guy comes back and confirms it or denys it etc then we cant really move on and if it was = they all have the same base so its just a quadratic and the lg the answers

Reply 10

steve2005

17

massimo

well i answered the question as written and i think its the right answer so far unless this guy comes back and confirms it or denys it etc then we cant really move on and if it was = they all have the same base so its just a quadratic and the lg the answers

I think your answer is incorrect. If we assume it is an expression then you can't cancel. If it was an equation then you are allowed to cancel.

eg

40x²+80x=40x(x+2)

40x²+80x?x(x+2)

Reply 11

kikzen

14

massimo

well i answered the question as written and i think its the right answer so far unless this guy comes back and confirms it or denys it etc then we cant really move on and if it was = they all have the same base so its just a quadratic and the lg the answers

you cant split up logs like that anyway (like how you did)

Reply 12

massimo

you cant split up logs like that anyway (like how you did)

well i thought you could and make log₁₀x

and then plug x into a quadratic and then just log₁₀ the answers

also he did state they are natural logs which is log₁₀ so i think i have got it right

lg(5x+8) + lg(2x+1) - 2lg(x+1)

but then if its an expression it would be

lg₁₀5x+lg₁₀8+lg₁₀2x+lg₁₀1-2lg₁₀x+2lg₁₀1

please correct if you feel its wrong thanks.

Reply 13

super_baros

massimo

lg(5x+8) + lg(2x+1) - 2lg(x+1)

but then if its an expression it would be

lg₁₀5x+lg₁₀8+lg₁₀2x+lg₁₀1-2lg₁₀x+2lg₁₀1

please correct if you feel its wrong thanks.

Hmmm.... I dont think you can go from lg(5x+8) = lg₁₀5x+lg₁₀8

I mean.. if lg is lg₁₀... then if you tappidy tap it out on the calculator...and take x as something like 2... then...

lg₁₀(5 x 2 +8) = 1.225
lg₁₀5 x 2 +lg₁₀8 = 1.903....

1.225 = 1.903???????

so yeh.. you can't do what you did...

Reply 14

Robob

17

massimo

i got lg5x+lg8+log2x+log1-log(x+1)²

log80x²÷logx²+2x+1=log(80+40x+80x²)

log(2x²+1+2)

so x=-0.25±√15÷16
log₁₀n=-0.25+√15÷16
log₁₀n=-0.25-√15÷16
and then just plug it in and get the answer but i need confirmation of this as im not sure its right !!

that's wrong lg(5x+8) ≠ lg5x + lg8

Reply 15

massimo

that adds another dimension is it supposed to be = 0 the answer to be honest i think its supposed to be - rahter than = but until this guy comes back we cant be for sure but then we go back into my quadratic lg thingys !!

Reply 16

super_baros

massimo

that adds another dimension is it supposed to be = 0 the answer to be honest i think its supposed to be - rahter than = but until this guy comes back we cant be for sure but then we go back into my quadratic lg thingys !!

The dude was just online..........

Can't he (or she) confirm what is was.... Can't help if the question is wrong....

Oh btw.. I don't have a clue what you did with your quadratic logs thingy etc.... if it is = 0 then you only need to solve the quadratic equation like steve2005 showed earlier....

Reply 17

Robob

17

lg(5x+8) + lg(2x+1) - 2lg(x+1)
= lg [(5x+8)(2x+1)] - lg (x+1)²
=lg[((5x+8)(2x+1))/(x+1)²]

Reply 18

Geerththanan

OP

17

I am so sorry for the late reply to those who asked me to reply, here is the link from where I got the question wrong, I just copied and pasted it, coz i could not figure it out myself...

http://nrich.maths.org/askedNRICH/edited/2274.html

:smile:

....once again, appologies for late responce...

Reply 19

Libertine

8

lgeerthan

I am so sorry for the late reply to those who asked me to reply, here is the link from where I got the question wrong, I just copied and pasted it, coz i could not figure it out myself...

http://nrich.maths.org/askedNRICH/edited/2274.html

:smile:

....once again, appologies for late responce...

Yep, the original question makes no sense.

Logarithms...

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