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Mathematics 14 Online
OpenStudy (jazzyfa30):

log 3 5 + log 3 x = log 3 10

OpenStudy (jazzyfa30):

no the threes are at the bottom of the log

hero (hero):

\[\log_3(5) + \log_3(x) = \log_3(10)\]

OpenStudy (jazzyfa30):

yes Thank you @Hero

hero (hero):

Remember \(\log(a) + \log(b) = \log(ab)\)

OpenStudy (anonymous):

log x + log (x+3) = log 10 log x(x+3) = log 10 x(x+3) = 10 x^2 + 3x = 10 x^2 + 3x - 10 = 0 (x-2)(x+5) = 0 x = 2 or x = -5

OpenStudy (jazzyfa30):

u think u smart @50_cent

OpenStudy (jdoe0001):

\(\bf \log_3(5) + \log_3(x) = \log_3(10)\\ log_3(5\times x) = \log_3(10)\\ \text{log cancellation rule of}\\ a^{log_ax} = x\\ 3^{log_3(5\times x)} = 3^{\log_3(10)} \implies 5x = 10\)

OpenStudy (anonymous):

im so smart

hero (hero):

@50_cent steps are not correct

OpenStudy (anonymous):

then show me hero

OpenStudy (jazzyfa30):

hahahaha but seriously hero how i do this junk

hero (hero):

The steps are pretty simple \[\log_3(5) + \log_3(x) = \log_3(10) \\ \log_3{(5x)}=\log_3(10) \\5x = 10\]

hero (hero):

Solve for x

OpenStudy (anonymous):

ohhhhhhhhhhhh forgot to put 5 x

hero (hero):

@50_cent, your steps are simply not correct.

hero (hero):

@jdoe0001 did extra steps that were unnecessary.

OpenStudy (anonymous):

ohh

OpenStudy (anonymous):

OMG @Hero

OpenStudy (jazzyfa30):

so x=2

hero (hero):

Correct. Don't forget to check.

OpenStudy (jazzyfa30):

ok can u help me with a few more

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