exponential question (1 Viewer)

[StGeorgeDragons]

solve for x:

log (base e) 3, to the power of 2x+5 = 1



(looks like log little e 3 to power of 2x + 5 = 1 )

 

[SaHbEeWaH]

recall that if xⁿ = 1 and x does not equal 0 or 1
n = 0
(well just think about it, don't need to recall it)
anything to the power of 0 = 1
we know that ln3 does not equal 0 or 1
so, n = 0
2x + 5 = 0
x = -5/2

 

[Antwan23q]

ln3^(2x+5)=1

3^(2x+5)=e^1

(2x+5)log3=log e

2x+5= log e / log 3

2x = log(base 3) e)-5

x=(log(base 3)e -5)/2

 

[FinalFantasy]

do u mean the whole thing to da power 2x+5 or 3 to the power of 2x+5

 

[Antwan23q]

recall that if xⁿ = 1 and x does not equal 0 or 1
n = 0
(well just think about it, don't need to recall it)
anything to the power of 0 = 1
we know that ln3 does not equal 0 or 1
so, n = 0
2x + 5 = 0
x = -5/2

Nah mate, that dont work, cause that gives u Log of base e 1
which equals zero. it doesnt equal one

 

[SaHbEeWaH]

solve for x:

log (base e) 3, to the power of 2x+5 = 1



(looks like log little e 3 to power of 2x + 5 = 1 )


course you could do it algebraically:

(ln3)^2x+5 = 1
take the logs of both sides
ln[(ln3)^2x+5] = ln1
2x+5 . <b>ln</b>(ln3) = ln1
2x+5 = 0
x = -5/2

edit: small mistake
although didn't make a difference since it was a constant but nonetheless

 

[FinalFantasy]

solve for x:

log (base e) 3, to the power of 2x+5 = 1



(looks like log little e 3 to power of 2x + 5 = 1 )


course you could do it algebraically:

ln3^2x+5 = 1
take the logs of both sides
ln[(ln3)^2x+5] = ln1
2x+5 . ln3 = ln1
2x+5 = ln1/ln3
2x+5 = 0
x = -5/2

ln 3^(2x+5)=1
if x=-5\2
LHS=ln 3^(-5+5)=ln 3^0=ln 1=0

 

[SaHbEeWaH]

solve for x:

log (base e) 3, to the power of 2x+5 = 1

he placed a comma after the three,
this means
(log_e3)^2x+5 = 1

not log_e(3^2x+5) = 1
else he'd say
log (base e), 3 to the power of 2x + 5 = 1

 

[FinalFantasy]

if its da whole thing to power of ..... den ur first solution is right

however there is a little error in da 2nd one

 

[FinalFantasy]

o ur doing da same thing except diff way, sorry i tot u doing it just 3^.... on the 2nd one LoL

 

[Antwan23q]

yeh, i can see that.

 

[FinalFantasy]

its hard to distinguish wat the question is when typing these log things here

 

[KFunk]

ln3^(2x+5)=1

3^(2x+5)=e^1

(2x+5)log3=log e

2x+5= log e / log 3

2x = log(base 3) e)-5

x=(log(base 3)e -5)/2

Just an alternative I'd like to offer. e^lna = a so it follows that:

log_e3^2x+5=1

3^2x+5= e

e^ln3(2x+5)= e¹

ln3(2x+5) = 1

2x +5 = 1/ln3

x = 1/2.ln3 - 5/2