Why Does the Equation log(35-x^3)/log(5-x)=3 Simplify to 1000=(35-x^3)/(5-x)?

  • Thread starter Thread starter thomasrules
  • Start date Start date
AI Thread Summary
The equation log(35-x^3)/log(5-x)=3 simplifies to 1000=(35-x^3)/(5-x) through the property of logarithms. By rewriting the logarithmic equation, it becomes clear that taking the anti-log leads to the relationship between the two expressions. The correct first step involves expressing the logarithmic equation in terms of natural logarithms, specifically ln(35-x^3) = 3 ln(5-x). This transformation allows for further simplification and solving of the equation. The discussion concludes with the participant confirming their understanding of the simplification process.
thomasrules
Messages
243
Reaction score
0
log(35-x^3)/log(5-x)=3

I can't get it...

What I did was: 1000=(35-x^3)/(5-x)
 
Physics news on Phys.org
Putting this in a better grouping you get 3log(5-x)=log(35-x^{3})

Take the anti-log of both sides from there.
 
Last edited by a moderator:
\frac {\ln x} {\ln y} \neq \ln ( \frac x y) so what you have is not correct.


Your first step should be

\ln (35- x ^3) = 3 \ln (5-x)

Can you proceed from there?
 
i got it thank you...
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top