0
$\begingroup$

My strategy for evaluating the logarithms of surds that are in the form of: log$_b$ ($^x\surd$rad) = y involves transforming the surd to exponent form, where the radicand becomes the base and the index is the reciprocal of the surds’ and putting the whole thing into what I think of as index notation (b$^x$= y)

For example, evaluating log$_3$ ($^4\surd$9) = x converts to log$_3$ (9$^{1/4}$) = x

In the next stage of the process, I rearrange the initial statement to the form b$^x$ = y and perform further operations on the Left Hand Side (LHS) to deduce x.

For example:
step 0: 3$^x$ = 9$^{1/4}$
step 1: (3$^2$)$^x$ = 9$^{1/4}$
step 2: (3$^2$)$^{1/4}$ = 9$^{1/4}$
step 3: (3$^{2/4}$) = 9$^{1/4}$
step 4: (3$^{1/2}$) = 9$^{1/4}$
… final answer: log$_3$ ($^4\surd$9) = 1/2

What’s got me confused is that I arrived at the process in the second half by a flash of insight/intuition and am struggling to determine how the operations I perform on the LHS are legal, since in step 1 I appear to change the value of one side without changing the value of the other side.

I can only conclude that my understanding of equation transposition must be flawed. Request for comment.

$\endgroup$
1
  • $\begingroup$ Cye Waldman's comment is definitely correct. $3^x = 9^{1/4}$. Taking the natural logarithm of each side gives $\ln 3^x = \ln 9^{1/4}$. Since $\ln b^t = t\ln b$ whenever $b > 0$ and $b \neq 1$, $x\ln 3 = \frac{1}{4}\ln 9$. Since $9 = 3^2$, $x\ln 3 = \frac{1}{4} \ln 3^2$. Using the rule $\ln b^t = t\ln b$ again, gives $x\ln 3 = \frac{1}{2}\ln 3$. Since $\ln 3 \neq 0$, we may divide both sides of the equation by $\ln 3$ to obtain $x = 1/2$. $\endgroup$
    – N. F. Taussig
    Apr 18, 2022 at 9:31

1 Answer 1

Reset to default
3
$\begingroup$

I think the confusion is in how you write the steps. In particular, I am confused by how you went from step 0 to 1 as well. Let me rephrase what I think you are trying to say, and see if it makes more sense.

Step 0: We want to solve $3^x = 9^{\frac{1}{4}}$.

Step 1: We know that $9 = 3^2$, so $\color{red}{\text{RHS}} = (3^2)^{\frac{1}{4}}$, meaning $3^x = (3^2)^{\frac{1}{4}}$.

Step 2: Now, we follow your step $3$ and $4$ to get $\color{red}{\text{RHS}} = 3^{\frac{1}{2}}$, meaning $3^x = 3^{\frac{1}{2}}$.

Finally, this means $x = \frac{1}{2}$.

Does this help?

$\endgroup$
1
  • $\begingroup$ Yes… this is it. Thank you Gareth. That’s consistent with my learning to date. The process of equating indexes requires that the bases be made similar first, as you do. $\endgroup$
    – duckegg
    Apr 17, 2022 at 23:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .