Logarithmic Challenge...please help! (log base 10 3)/10... will award 10 pts?

Question

The equation is (log base 10 3)/10 = [log base 10 (1+r)]

I've tried to use the distributive property but that DEFINITELY won't work because then log 1 + log r = log r, which is not log (1+r). So, can you use antilogs (anything with log) to solve and show your steps?

Thanks!

oops, it's actually (log base 10 120/49)/10

Answer 1

Hey there!

Log base 10 is the same as a common log, or log.

Here's the answer.

log(3)/10=log(1+r) --> Write the problem.
log(3)=10log(1+r) --> Multiply 10 to both sides of the equation.
log(3)=log((1+r)^10) --> Use the power property of logarithms on the right side of the equation i.e. use the fact that nlog(m)=log(m^n).
3=(1+r)^10 --> Use one-to-one property of logarithms i.e. if log(m)=log(n), then m=n.
3^(1/10)=1+r --> Take the tenth root on each side of the equation.
3^(1/10)-1=r --> Subtract 1 on both sides of the equation.
r=3^(1/10)-1 Use symmetric property of equality i.e. if a=b, then b=a.

So the answer is r=3^(1/10)-1 or r is about equal to 0.116.

Wait. In your additional details section, you said that it was (log base 10 120/49)/10.

Well here's the correct answer.

log(120/49)/10=log(1+r) --> Write the problem.
log(120/49)=10log(1+r) --> Multiply 10 to both sides of the equation.
log(120/49)=log((1+r)^10) --> Use the power property of logarithms on the right side of the equation i.e. use the fact that nlog(m)=log(m^n).
120/49=(1+r)^10 --> Use one-to-one property of logarithms i.e. if log(m)=log(n), then m=n.
120/49^(1/10)=1+r --> Take the tenth root on each side of the equation.
120/49^(1/10)-1=r --> Subtract 1 on both sides of the equation.
r=120/49^(1/10)-1 Use symmetric property of equality i.e. if a=b, then b=a.

So the correct answer is r=(120/49)^(1/10)-1, or r is approximately equal to 0.094.

Hope it helps!

Answer 2

I'll assume all logs are to base 10, to simplify the notation.

(log 120/49)/10 = log (1+r)
=> log ((120/49)^(1/10)) = log(1+r)
=> (120/49)^(1/10) = 1+r
=> r = (120/49)^(1/10) - 1
= 0.0937 to 4 d.p.

Answer 3

[ log_10 (120/49) ] / 10 = log_10 (1 + r)

[ log_10 (120) - log_10 (49) ] / 10 = log_10 (1 + r)

10^ [ [ log_10 (120) - log_10 (49) ] / 10 ] = 1 + r

r = 10^ [ [ log_10 (120) - log_10 (49) ] / 10 ] - 1