Question

logarithms! anyone willing to help meh? :)

Answer 1

Okie, da question is:
Solve the following logarithmic equation.
\(log_5(x+15) + log_5(x+35)=3\)

Answer 2

\[\log_b a+\log_b c = \log_b(ac)\]

Answer 3

\[\huge\rm \log_b (x) + \log_b y= \log_b (x\cdot y)\]
addition to multiplication \[\log_b x - \log_b y = \log_ b \frac{x}{y}\]
subtraction to division

Answer 4

So \(log_5((x+15)(x+25))= 3\)?

Answer 5

so you can multiple the interior of the 2 logarithms and then you can convert to exponential form

Answer 6

*x+35

Answer 7

yes correct

Answer 8

Okay, so doing that I get \(log_5(x^2 + 50x + 525) = 3\), what do I do from there?

Answer 9

/color green

Answer 10

/color yellow

Answer 11

Convert to exponential form is a possibility.

Answer 12

ah, forgot about that, x^2 + 50x + 525 = 3^5?

Answer 13

how did you get x^2+50x+525 ??

Answer 14

or would it be 5^3? can never remember which way...

Answer 15

is it +25 of +35?

Answer 16

+35

Answer 17

its actually opposite:
opposite side of log becomes exponent of the base \[\log_b x =y \rightarrow b^y =x\]

Answer 18

+35, I made a mistake the second time I typed it, but then corrected it

Answer 19

ooohhhh the base of the log is the base of the exponent

Answer 20

\[\log_5\left(x+15\right)+\log_5\left(x+35\right)=3\]
\[\mathrm{Apply\:\log\:rule}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)\]
\[\log _5\left(x+15\right)+\log _5\left(x+35\right)=\log _5\left(\left(x+15\right)\left(x+35\right)\right)\]
\[\log _5\left(\left(x+15\right)\left(x+35\right)\right)=3\]
\[\mathrm{Apply\:\log\:rule}:\quad \:a=\log _b\left(b^a\right)\]
\[3=\log _5\left(5^3\right)=\log _5\left(125\right)\]
\[\log _5\left(\left(x+15\right)\left(x+35\right)\right)=\log _5\left(125\right)\]
\[\mathrm{When\:the\:logs\:have\:the\:same\:base:\:\:}\log _b\left(f\left(x\right)\right)=\log _b\left(g\left(x\right)\right)\quad \Rightarrow \quad f\left(x\right)=g\left(x\right)\]
\[\mathrm{For\:}\log _5\left(\left(x+15\right)\left(x+35\right)\right)=\log _5\left(125\right)\mathrm{,\:\quad solve\:}\left(x+15\right)\left(x+35\right)=125\]
\[\left(x+15\right)\left(x+35\right)=125\]
\[\mathrm{Solve\:}\:\left(x+15\right)\left(x+35\right)=125:\quad x=-10,\:x=-40\]
\[\mathrm{Verifying\:Solutions}:\quad x=-10\space\mathrm{True},\:\spacex=-40\space\mathrm{False}\]

Answer 21

So final Answer is...
\[x=-10\]

Answer 22

Yes because you cant take the logarithm of a negative number so -40 is false

Answer 23

@sleepyjess are you gotit

Answer 24

Where did \(log_5(125)\) come from?

Answer 25

hes rewriting the number 3 for some reason.

Answer 26

Couldn't I just do x^2 + 50x + 525 = 125, then go x^2 + 50x + 400 = 0 and either factor or use quadratic formula from that?

Answer 27

you could do that and you would be perfectly fine doing that

Answer 28

more than one way to skin a cat

Answer 29

Okay, I'd rather just stick to the methods I know how to do than try to learn a different way the night before the final

Answer 30

I agree. The second way presented is very convoluted in my opinion.

Answer 31

Would you mind helping me with one more @sweetburger ?