Exponential Equation - Let’s solve the equation using logarithms

How to use logarithms to solve an exponential equation.
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[@joseeoliviero6078]

Don't need to use log, in this case, if you know your "rules of power". 3^2x+1 = 81. Change 81 to 3^4. (rule - if a^b=a^c; b=c). Therefore, 2x+1=4. 2x=3, x=3/2 or 1.5.

[@johnplong3644]

That is how I solve this problem

[@thewolfdoctor761]

Yes, it is obviously the easy method to solve for x. Anyone could solve it their head.

[@beachman8106]

Absolutely ❤

[@robertgamble7497]

I viewed your solution and when I solved it I didn’t distribute the log3 into (2x + 1). I got the same answer and it was easier:
Thank you for all the great teachings.
Math Problems
➖➖➖➖➖➖➖
3^(2x + 1) + 5 = 86
3^(2x + 1) = 81
log3^(2x + 1) = log81
(2x + 1)log3 = log81
(2x + 1) = log81/log3
(2x + 1) = 4
2x = 3
x = 3/2
➖➖➖➖➖➖➖➖

[@guyreece6425]

I like your step wise method of solving. Thanks for your efforts.

[@Dark1inside]

Thanks!

[@y0us3rn4m3]

This one is easy to solve without logarithms. It's easy to see that (2x+1) = 4), So 2x=3 and X=1.5 (exactly, not approximately).

[@deograciousuwiragiye8429]

86-5=81
81=3 power 4
Now can equalize
2x+1=4
2x =3
X =2/3
If I make proof by replacing x by its value,the equation is correct.

[@billrandle4437]

The equation can be rewritten
3^(2x+1) = 86 - 5
i. e. 3^(2x+1)=81
Now 81 =3^4
Equating exponents gives us
2x+1=4 that is x = 3/2

[@thatomofolo452]

Solid points though ✍️✍️✍️

[@johnplong3644]

No calculators back in my Day just tables in the back of the book I did this using the rules of exponents ( power rule)1.5

[@user-ri6rn7ti5h]

(x+3x-3) (x+1x-1)

[@kennethwright870]

2x+1 obviously equals 4, since 3^4=81, so x =3/2

[@deograciousuwiragiye8429]

X=3/2

[@TSR1942]

X=one

[@russophilej857]

3 raised to what power is 81 ??? Set
exponent equal to 4 ... it worked this
time ... but not the "general case" ...
Logs are the better way !!! Be well ...

[@murdock5537]

2x + 1 = 4 → x = 3/2 🙂

[@danielmadden9691]

X=1.5

[@aryusure1943]

I got it but most certainly not with the right method.
But common sense still works since it's easy to see that 3 to the power 2x + 1 = 81 (86 minus 5).
Once you get that you notice that 81 = 3 to the 4th power (3 X 3 X 3 X 3 = 81).
So 2x + 1 must equal 4.
So 2x = 3 (4 minus 1)
Bingo! X = 1.5

[@mracjesstark3468]

No need to use log..where 81=3⁴
2X+1-4=0
2x-3=0
2x=3
X=3/2
X=1.5

[@olivemd]

I got the correct answer, but your explanation confused me.
Edit: I will listen a few more times. Maybe I should ask my son if he can give me an old scientific calculator. I’m from the ancient times( before calculators).

[@jjfishingproductions1271]

Just get to the point man you talk to much just get to the point it’s wasting time and it’s annoying

[@homepropsllc2485]

Why do u confuse us with all these logs, f(x), lns, sins, cos, etc.
I got migraine just looking and listening to ur method.
I solved within 12 seconds as follows:
PROBLM.
"""”""""""""""""”"
3^(2x+1) + 5 = 86
SOLTN.
3^(2x+1) = 86 - 5
3^(2x+1) = 81
3^(2x+1) = 3⁴
Therefore, 2x+1= 4
2x=4-1
2x=3
x =3/2 =1.5 Ans.

[@ndailorw5079]

Yeah… and well done …and much quicker …and much easier…and correct, to boot! But this particular problem is pretty obvious and straightforward, it’s clean and smooth and rounded. We can easily see that 3 to some power (2x + 1) = 81, which itself is just 3 to the 4th power = 81. The the point of the instructor is to teach how to solve exponential functions in general by using their inverse logarithmic function, and vice versa. And as he says beginning @ 10:16 to about 10:56, sometimes the final answer may be required to be expressed In logarithmic notation rather than its final numerical form,
What happens when the problem isn’t quite as straightforward and obvious, say, like, 3^(2x + 1) - 7 = 19683? We could use your method and make a few trial and error but educated guesses , and may get lucky on the first try, but that would be time consuming as well as unnecessary and we’d be better off simply taking logs. But if we’re not versed and skilled in taking logs, then what! For this particular problem we can easily see that 3^(2x + 1) = 81 is the same thing as 3^(2x + 1) = 3^4 and it’s all a simple matter from there. But what about the problem I give above? A somewhat fundamental problem, but this time the exponent isn’t so obvious and clean. So that it’s best to take the log on both sides rather than guess at the exponent. But if we’re not certain in how to use logarithms we’re stuck and done for. Or if the teacher wants the final answer and the steps before it expressed in terms of logarithms rather than a number we’re screwed. That’s why this instructor is going about things the way he is here. I’m quite sure he can do the problem he’s doing here just as fast and as quick and as easy as you did here…he knows that. But that’s not his point and purpose here. But rather, his goal and purpose is to teach us how to solve exponential functions logarithmically, and logarithmic functions exponentially, since the two functions are inverses of each other! Your method is excellent! But the point is to be armed at all times with all methods so that the more needful and appropriate method may be applied. Can’t use your method on a problem like, 3^(2x + 1) - 4 ≈ 121.6994 …then what? The instructor’s goal and purpose here is not to show how to solve the problem quickly, he’s teaching how to solve problems that can’t be solved quickly …he’s teaching how to use logarithms! He knows how to do it the way you’ve done it here, but that’s not what he wants!

[@rossbatrossbdotcom]

He kept saying the answer is approximately 1.50 when it is in fact, exactly 1.50

[@bobwineland9936]

X=3/2