Exponential functions

[Astrea]

Hi,

I've been trying to solve this exponential function problem and I'm stuck.

6^x/2 = 4<sup>(1-x)


I took natural log of both sides to get:

(x/2) ln 6 = (1-x) ln 4

and end up with

(x)/ (2-2x) = ln 4 / ln 6

I am having trouble isolating the x after this step. I am not sure if something escapes me or I made a mistake before arriving to this part. Thank you.</sup>

 

[pka]

I've been trying to solve this exponential function problem and I'm stuck.
6^x/2 = 4^{(1-x)
I took natural log of both sides to get:
(x/2) ln 6 = (1-x) ln 4}

\(\displaystyle \begin{align*} \left( {\frac{x}{2}} \right)\ln (6) &= (1 - x)\ln (4)\\ x\ln(6)&=\ln(16)-x\ln(16)\\x &=\frac{\ln(16)}{\ln(96)}\end{align*}\)

 

[Astrea]

\(\displaystyle \begin{align*} \left( {\frac{x}{2}} \right)\ln (6) &= (1 - x)\ln (4)\\ x\ln(6)&=\ln(16)-x\ln(16)\\x &=\frac{\ln(16)}{\ln(96)}\end{align*}\)

Could you explain your thought process a bit on this? I am trying to see it but I just don't get it.

 

[pka]

Could you explain your thought process a bit on this? I am trying to see it but I just don't get it.

Do you see that \(\displaystyle 2\ln(4)=\ln(16)~? \)

If you do then can you solve \(\displaystyle xc=b-xb \) for \(\displaystyle x ~?\)

Let \(\displaystyle c=\ln(6)~\&~b=\ln(16)~.\)

Do you understand that \(\displaystyle \ln(6)+\ln(16)=\ln(96)~? \)

 

[Astrea]

Thank you. That log property escaped my brain.