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log[3](2x+5) = log[3]9+2
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\[\huge\rm log_3 (2x +5) = \log_3 (9x+2)\] i guess that suppose to be 9x right ??
right side
??
um no its not
so 9+2 ??
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yes
are u sure ?
thats what the paper says
\[\huge\rm log_3 (2x +5) = \log_3 (9+2)\] okay then both side have log_3 base so you can cancel out
ok \[2x+5 = 9+2\]
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yes solve for x
why do the logs cancel out ? my teacher hasn't told us anything abut that i dont believe
bec both side log r same \[\log_3 = \log_3\] what you would get when you divide both side by log_3 ?
1
yes so 1 (2x+5) = 1(9+2)
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oh ok
what times are you on here because you are a HUGE help !!
usually 3-9 pm
ok you will most likely find me on here tomorrow . and is that pacific time?
nope
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