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OpenStudy (anonymous):
solve the given equation
log5 (x+2) - log5 11= log5 121 ?
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OpenStudy (anonymous):
@jdoe0001
OpenStudy (anonymous):
can you show me @jdoe0001
OpenStudy (jdoe0001):
well, let us try using the 2nd rule listed there on the left hand side.... what would that give you?
OpenStudy (jdoe0001):
\(\bf log_5(x+2) - log_5(11)= log_5(121)\implies log_5\left(\cfrac{x+2}{11}\right)= log_5(121)
\\ \quad \\
\cfrac{x+2}{11}=121\)
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OpenStudy (anonymous):
so is that the answer @jdoe0001
OpenStudy (jdoe0001):
you need to solve for "x"
OpenStudy (anonymous):
how would we do that @jdoe0001
OpenStudy (anonymous):
@jdoe0001 ??
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OpenStudy (anonymous):
@surjithayer ??
OpenStudy (anonymous):
x+2=121*11=?
find x
OpenStudy (bmorse):
now multiply: 121*11= x+2
1331= x+2
x= 1331-2 = 1329
OpenStudy (anonymous):
are you sure this is correct @bmorse
OpenStudy (bmorse):
hang on let me check...
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OpenStudy (bmorse):
Yes it is!!
OpenStudy (anonymous):
thanks
OpenStudy (bmorse):
you're welcome!
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