Solving for a variable in a complex algebra equation with logs and powers

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1
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I have a follow-up question to my last one. I need to solve for b for the following:



$$d=fracs(ln(o))^bs(ln(g))^b$$



Once again, this is beyond the level of several online algebra calculators.




algebra-precalculus logarithms exponentiation




share|cite|improve this question




















  • Why do you need an online calculator? The RHS is just the ratio raised to the power of $b$.
    – NickD
    Aug 11 at 22:56










  • Well, true. It's still a complex process to determine b = ...
    – IamIC
    Aug 11 at 22:57










  • Nothing complex about it.
    – NickD
    Aug 11 at 22:58










  • For someone well versed in this math, of course it's simple. But, this isn't my line. Please feel free to answer the question.
    – IamIC
    Aug 11 at 22:59






  • 2




    If your equation was $d = a^b$ can you solve for b?
    – NickD
    Aug 11 at 23:01














up vote
1
down vote

favorite












I have a follow-up question to my last one. I need to solve for b for the following:



$$d=fracs(ln(o))^bs(ln(g))^b$$



Once again, this is beyond the level of several online algebra calculators.




algebra-precalculus logarithms exponentiation




share|cite|improve this question




















  • Why do you need an online calculator? The RHS is just the ratio raised to the power of $b$.
    – NickD
    Aug 11 at 22:56










  • Well, true. It's still a complex process to determine b = ...
    – IamIC
    Aug 11 at 22:57










  • Nothing complex about it.
    – NickD
    Aug 11 at 22:58










  • For someone well versed in this math, of course it's simple. But, this isn't my line. Please feel free to answer the question.
    – IamIC
    Aug 11 at 22:59






  • 2




    If your equation was $d = a^b$ can you solve for b?
    – NickD
    Aug 11 at 23:01












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I have a follow-up question to my last one. I need to solve for b for the following:



$$d=fracs(ln(o))^bs(ln(g))^b$$



Once again, this is beyond the level of several online algebra calculators.




algebra-precalculus logarithms exponentiation




share|cite|improve this question












I have a follow-up question to my last one. I need to solve for b for the following:



$$d=fracs(ln(o))^bs(ln(g))^b$$



Once again, this is beyond the level of several online algebra calculators.





algebra-precalculus logarithms exponentiation





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 11 at 22:51









IamIC

3211215




3211215











  • Why do you need an online calculator? The RHS is just the ratio raised to the power of $b$.
    – NickD
    Aug 11 at 22:56










  • Well, true. It's still a complex process to determine b = ...
    – IamIC
    Aug 11 at 22:57










  • Nothing complex about it.
    – NickD
    Aug 11 at 22:58










  • For someone well versed in this math, of course it's simple. But, this isn't my line. Please feel free to answer the question.
    – IamIC
    Aug 11 at 22:59






  • 2




    If your equation was $d = a^b$ can you solve for b?
    – NickD
    Aug 11 at 23:01
















  • Why do you need an online calculator? The RHS is just the ratio raised to the power of $b$.
    – NickD
    Aug 11 at 22:56










  • Well, true. It's still a complex process to determine b = ...
    – IamIC
    Aug 11 at 22:57










  • Nothing complex about it.
    – NickD
    Aug 11 at 22:58










  • For someone well versed in this math, of course it's simple. But, this isn't my line. Please feel free to answer the question.
    – IamIC
    Aug 11 at 22:59






  • 2




    If your equation was $d = a^b$ can you solve for b?
    – NickD
    Aug 11 at 23:01















Why do you need an online calculator? The RHS is just the ratio raised to the power of $b$.
– NickD
Aug 11 at 22:56




Why do you need an online calculator? The RHS is just the ratio raised to the power of $b$.
– NickD
Aug 11 at 22:56












Well, true. It's still a complex process to determine b = ...
– IamIC
Aug 11 at 22:57




Well, true. It's still a complex process to determine b = ...
– IamIC
Aug 11 at 22:57












Nothing complex about it.
– NickD
Aug 11 at 22:58




Nothing complex about it.
– NickD
Aug 11 at 22:58












For someone well versed in this math, of course it's simple. But, this isn't my line. Please feel free to answer the question.
– IamIC
Aug 11 at 22:59




For someone well versed in this math, of course it's simple. But, this isn't my line. Please feel free to answer the question.
– IamIC
Aug 11 at 22:59




2




2




If your equation was $d = a^b$ can you solve for b?
– NickD
Aug 11 at 23:01




If your equation was $d = a^b$ can you solve for b?
– NickD
Aug 11 at 23:01










2 Answers
2






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up vote
2
down vote



accepted










$$d=fracs(ln(o))^bs(ln(g))^b$$
$$d= left (fracln(o)ln(g) right )^b $$
Taking logs of both sides ...
$$ ln(d)=b ln left (fracln(o)ln(g) right ) $$



$$ b = frac ln(d) ln(ln(o))-ln(ln(g)) $$






share|cite|improve this answer




















  • Perfect, thank you!
    – IamIC
    Aug 11 at 23:07

















up vote
1
down vote













I interpreted $s(...)$ as being a function of one argument, as opposed to @WW1's answer where he interpreted it as a number multiplying both the numerator and the denominator of the fraction that could therefore be canceled. If my interpretation is correct, then



$$ d = left(fracs(ln(o))s(ln(g))right)^b$$



Taking logs, we then get



$$ b = fracln(d)ln(s(ln(o)) - ln(s(ln(g)) $$



which cannot be simplified further.






share|cite|improve this answer




















  • Thanks, NickD. @WW1's answer is correct.
    – IamIC
    Aug 14 at 9:06










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










$$d=fracs(ln(o))^bs(ln(g))^b$$
$$d= left (fracln(o)ln(g) right )^b $$
Taking logs of both sides ...
$$ ln(d)=b ln left (fracln(o)ln(g) right ) $$



$$ b = frac ln(d) ln(ln(o))-ln(ln(g)) $$






share|cite|improve this answer




















  • Perfect, thank you!
    – IamIC
    Aug 11 at 23:07














up vote
2
down vote



accepted










$$d=fracs(ln(o))^bs(ln(g))^b$$
$$d= left (fracln(o)ln(g) right )^b $$
Taking logs of both sides ...
$$ ln(d)=b ln left (fracln(o)ln(g) right ) $$



$$ b = frac ln(d) ln(ln(o))-ln(ln(g)) $$






share|cite|improve this answer




















  • Perfect, thank you!
    – IamIC
    Aug 11 at 23:07












up vote
2
down vote



accepted







up vote
2
down vote



accepted






$$d=fracs(ln(o))^bs(ln(g))^b$$
$$d= left (fracln(o)ln(g) right )^b $$
Taking logs of both sides ...
$$ ln(d)=b ln left (fracln(o)ln(g) right ) $$



$$ b = frac ln(d) ln(ln(o))-ln(ln(g)) $$






share|cite|improve this answer












$$d=fracs(ln(o))^bs(ln(g))^b$$
$$d= left (fracln(o)ln(g) right )^b $$
Taking logs of both sides ...
$$ ln(d)=b ln left (fracln(o)ln(g) right ) $$



$$ b = frac ln(d) ln(ln(o))-ln(ln(g)) $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 11 at 23:04









WW1

6,4821712




6,4821712











  • Perfect, thank you!
    – IamIC
    Aug 11 at 23:07
















  • Perfect, thank you!
    – IamIC
    Aug 11 at 23:07















Perfect, thank you!
– IamIC
Aug 11 at 23:07




Perfect, thank you!
– IamIC
Aug 11 at 23:07










up vote
1
down vote













I interpreted $s(...)$ as being a function of one argument, as opposed to @WW1's answer where he interpreted it as a number multiplying both the numerator and the denominator of the fraction that could therefore be canceled. If my interpretation is correct, then



$$ d = left(fracs(ln(o))s(ln(g))right)^b$$



Taking logs, we then get



$$ b = fracln(d)ln(s(ln(o)) - ln(s(ln(g)) $$



which cannot be simplified further.






share|cite|improve this answer




















  • Thanks, NickD. @WW1's answer is correct.
    – IamIC
    Aug 14 at 9:06














up vote
1
down vote













I interpreted $s(...)$ as being a function of one argument, as opposed to @WW1's answer where he interpreted it as a number multiplying both the numerator and the denominator of the fraction that could therefore be canceled. If my interpretation is correct, then



$$ d = left(fracs(ln(o))s(ln(g))right)^b$$



Taking logs, we then get



$$ b = fracln(d)ln(s(ln(o)) - ln(s(ln(g)) $$



which cannot be simplified further.






share|cite|improve this answer




















  • Thanks, NickD. @WW1's answer is correct.
    – IamIC
    Aug 14 at 9:06












up vote
1
down vote










up vote
1
down vote









I interpreted $s(...)$ as being a function of one argument, as opposed to @WW1's answer where he interpreted it as a number multiplying both the numerator and the denominator of the fraction that could therefore be canceled. If my interpretation is correct, then



$$ d = left(fracs(ln(o))s(ln(g))right)^b$$



Taking logs, we then get



$$ b = fracln(d)ln(s(ln(o)) - ln(s(ln(g)) $$



which cannot be simplified further.






share|cite|improve this answer












I interpreted $s(...)$ as being a function of one argument, as opposed to @WW1's answer where he interpreted it as a number multiplying both the numerator and the denominator of the fraction that could therefore be canceled. If my interpretation is correct, then



$$ d = left(fracs(ln(o))s(ln(g))right)^b$$



Taking logs, we then get



$$ b = fracln(d)ln(s(ln(o)) - ln(s(ln(g)) $$



which cannot be simplified further.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 12 at 11:57









NickD

9201412




9201412











  • Thanks, NickD. @WW1's answer is correct.
    – IamIC
    Aug 14 at 9:06
















  • Thanks, NickD. @WW1's answer is correct.
    – IamIC
    Aug 14 at 9:06















Thanks, NickD. @WW1's answer is correct.
– IamIC
Aug 14 at 9:06




Thanks, NickD. @WW1's answer is correct.
– IamIC
Aug 14 at 9:06












 

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