SOLUTION: I must solve the following equation, and then round the answer to the nearest hundreth, while using the Log way. 1+8(0.43)^2u=3
I placed Log in front of both equations like 2uLog
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Question 70761This question is from textbook Algebra 2 Explorations and Applications
: I must solve the following equation, and then round the answer to the nearest hundreth, while using the Log way. 1+8(0.43)^2u=3
I placed Log in front of both equations like 2uLog(0.43)=Log(3/9). However I keep getting a completely different answer than what the book has provided, which is .82. Please solve this solution, thanks.
This question is from textbook Algebra 2 Explorations and Applications
Found 2 solutions by Earlsdon, stanbon:
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Solve for u:
Subtract 1 from both sides of the equation.
Next, divide both sides by 8.
Simplify.
Now apply the logarithms.
Apply the power rule for logarithms to the left side.
Divide both sides by
Now divide both sides by 2.
Evaluate the logs using your calculator or log tables.
To the nearest hundredth.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
1+8(0.43)^2u=3
Subtract 1 from both sides to get:
8(0.43)^2u=2
log[8(0.43)^2u] =log2
log 8 + log(0.43)^2u =log 2
2u[log0.43] = log 2 - log8
2u[log0.43] = log(1/4)
2u = log(1/4)/log(0.43)
2u = 1.64258
u=0.82
--------
Cheers,
Stan H.
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