Question

logx+log(x+4)=5, the answer is 314.23409050891. Typically I would solve something like this log(x(x+4))=5; logx^2+4x=5 factor etc. But I am stuck on this one, and I don't know how to solve it.

Answer 1

log x = b

means

x = 10^b

Answer 2

log(x(x+4))=5

means

x(x+5) = 10^5

Answer 3

its a quadratic equation,
use quadratic formula... and u wil have to use calculator..

Answer 4

what base of the logarithm, base 10, base e, or something else?

Answer 5

if it is base 10

logx+log(x+4)=5
logx(x+4)=log5
x(x+4)=10^5
x^2+4x-10^5=0

which is your quadratic

Answer 6

It is base 10 if one of the solutions is \(-2+2\sqrt{25001}\approx 314.23409050891398163\)

Answer 7

yah when there is no base specified, the base is 10
10 is the base for all orphaned log expressions.. ;)

Answer 8

(note the quadratic will give you two solution, but one of them won't fit the original equation, [log of a negative won't be a real number])

Answer 9

It didn't specify a base, which also confused me. But, as you put, if it's ln, (base 10), that gives the correct solution. For some reason I forgot about the quadratic formula! Thanks guys! That helped a lot.

Answer 10

@ganeshie8 well, depends on the context. Log[x] in Mathematica is the natural logarithm, Log[10,x] is \(\log_{10}\), and it seems like I've encountered other systems that were like that as well. WolframAlpha also interprets "log" as "ln" unless you tell it otherwise. Traps for the unwary :-)

Answer 11

thats the basic problem that exists with all conventions i believe :)
however, by definition, from high school textbooks : log is base 10 and ln is base e

the programmer who designed matlab/mathematica did not honor the conventions. :P

Answer 12

whoops typo * logx(x+4)=log10^5

Answer 13

good to know! They switched the curriculum at my community college, so I wasn't taught anything about logarithms, and now, in business pre calculus they assume you have a basis of understanding, the instructor is doing a bit of review, and it is NOT review for me... So all of this information is incredibly helpful. Hopefully I'll be able to get myself up to speed. Thanks so much :)

Answer 14

fast way to review logarithms is to compare them wid exponent functions

Answer 15

and review inverse functions also..

Answer 16

Agreed that high school textbooks are almost certainly going to equate \(\log\) with \(\log_{10}\). However, different progenitors for MATLAB and Mathematica, so if they both went down that path (I haven't used MATLAB in decades, and don't recall), it appears that multiple people thought the natural log was the natural default :-)

Answer 17

Ahh yess... wolfram also thinks natural log is more natural so it interprets log as base e by default lol

Answer 18

but it does helpfully offer to interpret it otherwise most of the time.

Answer 19

review all below at the same time... they're all related closely :
1) log functions
2) exponent functions
3) inverse functions

Answer 20

I've often wondered why they aren't more typically taught together. Even people who seem to understand exponents well are often befuddled by logs being taught at another time, and inverse functions, well...

Answer 21

I'm getting that close relationship between logs & exponents. If anyone is reading this question feeling like I've been feeling this video really helped for me http://www.youtube.com/watch?v=r0YkfZxEaIc might help someone else with a similar lack of understanding.

Answer 22

nice video, ty for sharing :) u wil need to practice more... and stare at graphs of \(y=10^x \)and \(y = \log_{10} x\) etc.. a hundred times before feeling confident about logarithms and stuff.... good luck !

Answer 23

if \((a,b)\) is a point on graph \(y=10^x\),
then \((b, a)\) will exist on graph \(y = \log_{10} x\)