Finding values of x for logarithm The question is to find the numbers of x which satisfy the equation. log_x 10=log_4 100.

braffter92 2022-10-03 Answered
Finding values of x for logarithm
The question is to find the numbers of x which satisfy the equation.
log x 10 = log 4 100.
I have
ln 10 ln x = ln 100 ln 4 ln 10 ln x = 2 ln 10 2 ln 2
What would I do after this step?
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

Kaleb Harrell
Answered 2022-10-04 Author has 14 answers
All right. First multiply by l n ( x ) and by l n ( 2 ). You get
l n ( 10 ) l n ( 2 ) = l n ( 10 ) l n ( x )
Now divide by l n ( 10 ). This gives you
l n ( 2 ) = l n ( x )
Now you apply the exponential function on both sides to get rid of the logarithm:
e l n ( x ) = x = e l n ( 2 ) = 2
So x = 2
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-04-03
State domain, range, vertical asymptote and x,y intercepts for log4(x1)+1
asked 2022-07-16
Evaluate lim x 1 x 1 x ln ( 1 + 1 t ) d t
If the limit exists with l'Hopital i get
lim x ln ( 1 + 1 x ) 1 2 x = 2
but why can I apply l'Hopital, I mean does the integral 1 x ln ( 1 + 1 t ) d t really diverge ?
How can I show that ?
asked 2022-06-13
Solving for x in a equation involving natural logarithms
How would I solve for x in this equation here:
ln ( x ) + ln ( 1 / x + 1 ) = 3
I realize that the answer is e 3 1 , but I am not sure as to how to get it. Any input is appreciated.
asked 2022-09-27
Conditions required for ( z 1 z 2 ) ω = z 1 ω z 2 ω , where z 1 , z 2 , ω C
I am having trouble finding the conditions on z 1 and z 2 in order for:
( z 1 z 2 ) ω z 1 ω z 2 ω ω C
My first step was to rewrite the equation as:
e ω Log ( z 1 z 2 ) = e ω Log ( z 1 ) e ω Log ( z 2 )
We have that:
Log ( z 1 z 2 ) Log ( z 1 ) + Log ( z 2 ) + 2 n i π , n Z
And therefore, equating both sides of the equation, we get:
Log ( z 1 ) + Log ( z 2 ) + 2 n i π = Log ( z 1 ) + Log ( z 2 )
However, this appears to me to be true z 1 , z 2 C , yet if we set z 1 = z 2 = 1 and ω = i, then the equality does not hold, so I'm not sure where I'm making my error.
I believe that the condition required is when ( ( z ) R ( z ) = 0 ) ( z ) 0 because this is the point at which Log ( z ) is discontinuous, however, I am unable to prove that this is the case.
asked 2021-08-19
Please, write the logarithm as a ratio of common logarithms and natural logarithms.
log5(86)
a) common logarithms
b) natural logarithms
asked 2022-09-24
Integral 0 1 ln x ln ( 1 z x ) 1 x d x
How can I evaluate following logarithmic integral:
0 1 ln x ln ( 1 z x ) 1 x d x
asked 2022-07-19
problem solving logarithmic equation and reaching an equivalence
ok so i've had a problem trying to simplify the ln [ 1 + u 2 a 2 + u a ] and this is supposed to be equal to : ln [ a 2 + u 2 + u ]
how is this posible ?? i've tried to solve this for more than 2 hours and couldn't get to this equivalence. any suggestions ?