-1
$\begingroup$

How to solve this?

I am new to logarithms.

$$ \ x^{3\log^3 x-\big(\frac{2}{3}\big)\log x} = 100 \sqrt[3]{10}\ $$ All the logs have base $10$.

$\endgroup$
  • $\begingroup$ try taking the logarithm of both sides, isolate $\log(x)$$ and then raise it to an exponent 10. $\endgroup$ – Oria Gruber Jun 28 '17 at 15:50
  • $\begingroup$ I find it hard to believe that you are new to logarithms and yet you are given a question like this... $\endgroup$ – Feeds Feb 3 '18 at 7:18
2
$\begingroup$

It's $$\ (3\log^3{x}-\frac{2}{3}\log{x})\log{x} = 2\frac{1}{3}\ $$ and $\log{x}=t$

We get $$9t^4-2t^2-7=0,$$ which gives $t=1$ or $t=-1$ and the answer is: $$\left\{\frac{1}{10},10\right\}$$

$\endgroup$
  • $\begingroup$ Let me solve.... $\endgroup$ – Ravi Prakash Jun 28 '17 at 15:54
  • $\begingroup$ i have got two solutions $\endgroup$ – Dr. Sonnhard Graubner Jun 28 '17 at 16:01
  • $\begingroup$ for this error i got a nice inequality from you Michael $\endgroup$ – Dr. Sonnhard Graubner Jun 28 '17 at 16:04
  • $\begingroup$ I am sorry, i had mistakenly left x and cube. $\endgroup$ – Ravi Prakash Jun 28 '17 at 16:04
  • $\begingroup$ @Ravi Prakash I fixed my post for you. See now. $\endgroup$ – Michael Rozenberg Jun 28 '17 at 16:11
2
$\begingroup$

$3log(3)+ (2/3)log(x)= log(27+ x^{2/3}$ so $x^{3log(3)+ (2/3)log(x)}= x^{log(27+ x^{2/3})}$. Further, $100\sqrt[3]{10}= 10^{2+ 1/3}= 10^{7/3}$. So the equation becomes $x^{log(27+ x^{2/3})}= 10^{7/3}$ and taking the logarithm of both sides $(27+ x^{2/3})log(x)= 7/3$

$\endgroup$
  • $\begingroup$ @user247327 Your solution is total wrong! $3\log^3x\neq27$. $\endgroup$ – Michael Rozenberg Jun 28 '17 at 16:12
  • $\begingroup$ ?? I didn't say it was. $\endgroup$ – user247327 Jun 28 '17 at 16:27
1
$\begingroup$

Note that $$\log(100\sqrt[3]{10})=\log(100)+\log(\sqrt[3]{10})$$ and your equation $$(3\lg(x)^3-\frac{2}{3}\lg(x))\lg(x)=\lg(100)+\frac{1}{3}\lg(10)$$

$\endgroup$
  • $\begingroup$ Yes, its a nice explanation .....As the RHS upon simplification gives 2 + ⅓ $\endgroup$ – Ravi Prakash Jun 28 '17 at 16:40
  • $\begingroup$ What is $\text{lg}$? $\endgroup$ – Feeds Feb 3 '18 at 7:19
  • $\begingroup$ logarithm to the base $10$ $\endgroup$ – Dr. Sonnhard Graubner Feb 3 '18 at 7:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.