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Mathematics 9 Online
OpenStudy (anonymous):

Solve for x. Type your answer as an integer, like this: 13 log25x – log5 = 2

OpenStudy (jdoe0001):

http://www.teachengineering.org/collection/van_/lessons/van_bmd_less2/log_properties_lesson2_fig2.jpg look at the product property can you rewrite the left-hand-side?

OpenStudy (jdoe0001):

ahemm, rather look at the quotient property can you rewrite the left-hand-side?

OpenStudy (anonymous):

so basically its the opposite of subtraction which is your suppose to divide?

OpenStudy (jdoe0001):

the internal values, yes

OpenStudy (kirbykirby):

is that \[\log_{25}x-\log_{5}x ??\]

OpenStudy (anonymous):

25 divided by 5 = 5

OpenStudy (anonymous):

yes @kirbykirby

OpenStudy (jdoe0001):

right so log(25x/5) = log(5x) so log(5x) = 2 now exponentialize both the base of the log, or use the cancellation rules as $$ log(5x) = 2\\ \text{using cancellation rules}\\ 10^{log(5x)} = 10^2\\ 5x = 10^2 $$

OpenStudy (anonymous):

the answer is 2

OpenStudy (anonymous):

@jdoe0001

OpenStudy (jdoe0001):

bear in mind all I did for the cancellation rule is $$\large { log_\color{red}{{10}}(5x) = 2\\ \text{using cancellation rules}\\ \color{red}{10}^{log_{\color{red}{10}}(5x)} = \color{red}{10}^2\\ 5x = 10^2 } $$

OpenStudy (anonymous):

10 ^2=100/5=50

OpenStudy (jdoe0001):

100/5 = 50?

OpenStudy (anonymous):

sorry lol 20

OpenStudy (jdoe0001):

hehe

OpenStudy (anonymous):

would you mind elping me with a few more problems @jdoe0001

OpenStudy (jdoe0001):

sure, just post them in the channel, so we can all help and revise each other :)

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