SOLUTION: im not sure how to solve logarithmic and/or exponential equations! here is on of my questions... maybe some steps on how to find the answer would really help me! {{{ log7(x+1)=l

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: im not sure how to solve logarithmic and/or exponential equations! here is on of my questions... maybe some steps on how to find the answer would really help me! {{{ log7(x+1)=l      Log On

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Question 536640: im not sure how to solve logarithmic and/or exponential equations! here is on of my questions... maybe some steps on how to find the answer would really help me!
+log7%28x%2B1%29=log7%282xsquared-x-3%29+
Found 2 solutions by solver91311, josmiceli:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Equal logs with equal bases must have equal arguments.



So:



Collect terms and solve the resulting quadratic equation. Note: It won't factor over the rationals, so use the quadratic formula.

John

My calculator said it, I believe it, that settles it
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Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Some general rules are
+log%28a%29+%2B+log%28b%29+=+log%28a%2Ab%29+
+log%28a%29+-+log%28b%29+=+log%28a%2Fb%29+
+log%28a%5Eb%29+=+b%2Alog%28a%29+
You can use the rules in both directions
+log%287%2Cx%2B1%29=log%287%2C2x%5E2+-+x+-+3%29+
The rule to apply here is:
If this is true: +log%28a%2Cb%29+=+log%28a%2Cc%29+,
then +b+=+c+, so
+x+%2B+1+=+2x%5E2+-+x+-+3+
+x+%2B+1+=+%282x+-+3%29%2A%28x+%2B+1%29+ (I used trial and error)
Divide both sides by +x+%2B+1+
+1+=+2x+-+3+
+2x+=+4+
+x+=+2+
check answer:
+log%287%2C2%2B1%29=log%287%2C2%2A2%5E2+-+2+-+3%29+
+log%287%2C3%29+=+log%287%2C+8+-+2+-+3%29+
+log%287%2C3%29+=+log%287%2C+3%29+
OK