Exponentail Functions

[zbest1966]

5 *18^6x = 26
How did get that answer
Answer is x=-In(5) - In(26)/6(In(2)-2In3)

 

[jonah2.0]

DISCLAIMER: Beer soaked rambling/opinion/observation/reckoning ahead. Read at your own risk. Not to be taken seriously. In no event shall the wandering math knight-errant Sir jonah in his inebriated state be liable to anyone for special, collateral, incidental, or consequential damages in connection with or arising out of the use of his beer (and tequila) powered views.

5 *18^6x = 26
How did get that answer
Answer is x=-In(5) - In(26)/6(In(2)-2In3)

Are you sure?

 

[Ishuda]

5 *18^6x = 26
How did get that answer
Answer is x=-In(5) - In(26)/6(In(2)-2In3)

It looks like, in part, you have gotten you rules about logarithms mixed up:
ln(a b) = ln(a) + ln(b)
ln(a/b) = ln(a) - ln(b)
ln(a^b) = b ln(a)

 

[pka]

5 *18^6x = 26
How did get that answer x=-In(5) - In(26)/6(In(2)-2In3)

\(\displaystyle 5(18^{6x})=26\\\log(5)+6x[2\log(3)+\log(2)]=\log(13)+\log(2)\\x=\dfrac{\log(13)+\log(2)-\log(5)}{6[2\log(3)+\log(2)]}\)

 

[Steven G]

5 *18^6x = 26

Why not simply: x = LOG(26/5) / [6LOG(18)] ?

Like 2 * 6 = 12, not necessary to show 2 * 6 = 2 * 2 * 3 = 12

I must be not seeing something as I agree with Denis' result.

 

[Steven G]

5 *18^6x = 26
How did get that answer
Answer is x=-In(5) - In(26)/6(In(2)-2In3)

For the record it is Ln, not In. Ln stands for something like log natural.

 

[jonah2.0]

DISCLAIMER: Beer soaked rambling/opinion/observation/reckoning ahead. Read at your own risk. Not to be taken seriously. In no event shall the wandering math knight-errant Sir jonah in his inebriated state be liable to anyone for special, collateral, incidental, or consequential damages in connection with or arising out of the use of his beer (and tequila) powered views.

\(\displaystyle 5(18^{6x})=26\\\log(5)+6x[2\log(3)+\log(2)]=\log(13)+\log(2)\\x=\dfrac{\log(13)+\log(2)-\log(5)}{6[2\log(3)+\log(2)]}\)

Possible typo (or else I'm already too drunk).
This could have been
\(\displaystyle 5(18^{6x})=26\\\log(5)+6x[\log (2)+\log(3)+\log(3)]=\log(13)+\log(2)\\x=\dfrac{\log(13)+\log(2)-\log(5)}{6[\log(2)+\log(3)+\log(3)]}\)

 

[pka]

DISCLAIMER: Beer soaked rambling/opinion/observation/reckoning ahead. Read at your own risk. Not to be taken seriously. In no event shall the wandering math knight-errant Sir jonah in his inebriated state be liable to anyone for special, collateral, incidental, or consequential damages in connection with or arising out of the use of his beer (and tequila) powered views.

Possible typo (or else I'm already too drunk).
This could have been
\(\displaystyle 5(18^{6x})=26\\\log(5)+6x[\log (2)+\log(3)+\log(3)]=\log(13)+\log(2)\\x=\dfrac{\log(13)+\log(2)-\log(5)}{6[\log(2)+\log(3)+\log(3)]}\)

\(\displaystyle {6[\log(2)+\log(3)+\log(3)]}={6[\log(2)+2\log(3)]}\)

 

[jonah2.0]

DISCLAIMER: Beer soaked rambling/opinion/observation/reckoning ahead. Read at your own risk. Not to be taken seriously. In no event shall the wandering math knight-errant Sir jonah in his inebriated state be liable to anyone for special, collateral, incidental, or consequential damages in connection with or arising out of the use of his beer (and tequila) powered views.

\(\displaystyle {6[\log(2)+\log(3)+\log(3)]}={6[\log(2)+2\log(3)]}\)

Sorry about that Sir pka.
I was under the impression last night that the denominator was
6[log(3) + log(2)]
I didn't notice your 2 in 2log(3), leading my absinthe soaked brain to think that it may have been a typo.
No worries, happy to know I'm wrong; hard to feel down with my good friend, Lady Absinthe.
I've been noticing more errors on my reckonings the last few days though.
Maybe there's really something to what Sir Denis said about the insanity side effect of absinthe.
Maybe it's time to switch back to me old buddy tequila just to stay focused.
To keep riding the happiness wave with me mistress absinthe or to be focused with me old friend tequila. Hard stuff.

Cheers.

 

[lookagain]

I suggest leaving the answer with the original constants intact:

\(\displaystyle 5(18^{6x}) \ = \ 26\)

\(\displaystyle 18^{6x} \ = \ 26/5\)

\(\displaystyle log(18^{6x}) \ = \ log(26/5)\)

\(\displaystyle (6x)log(18) \ = \ log(26) \ - \ log(5)\)

\(\displaystyle x \ = \ \boxed{ \ \dfrac{log(26) \ - \ log(5)}{6*log(18)} \ }\)