Solve for x. Type your answer as an integer, like this: 13 log25x – log5 = 2
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?
ahemm, rather look at the quotient property can you rewrite the left-hand-side?
so basically its the opposite of subtraction which is your suppose to divide?
the internal values, yes
is that \[\log_{25}x-\log_{5}x ??\]
25 divided by 5 = 5
yes @kirbykirby
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 $$ log(5x) = 2\\ \text{using cancellation rules}\\ 10^{log(5x)} = 10^2\\ 5x = 10^2 $$
the answer is 2
@jdoe0001
bear in mind all I did for the cancellation rule is $$\large { log_\color{red}{{10}}(5x) = 2\\ \text{using cancellation rules}\\ \color{red}{10}^{log_{\color{red}{10}}(5x)} = \color{red}{10}^2\\ 5x = 10^2 } $$
10 ^2=100/5=50
100/5 = 50?
sorry lol 20
hehe
would you mind elping me with a few more problems @jdoe0001
sure, just post them in the channel, so we can all help and revise each other :)
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