MHB Solve for x: 4^(x-5) = 7^(2x-1) | Equation Help

  • Thread starter Thread starter Nikolas7
  • Start date Start date
AI Thread Summary
To solve the equation 4^(x-5) = 7^(2x-1), logarithms can be applied to both sides, leading to (x-5)log 4 = (2x-1)log 7. This transformation utilizes the property of logarithms that states log(a^x) = x log(a). The equation can then be manipulated as a linear equation to find the value of x. The discussion concludes with the user confirming their understanding of the logarithmic property used in the solution.
Nikolas7
Messages
22
Reaction score
0
Help me with the following equation:
${4}^{x-5}$=${7}^{2x-1}$
 
Mathematics news on Phys.org
Nikolas7 said:
Help me with the following equation:
${4}^{x-5}$=${7}^{2x-1}$

you can take log on both sides and get $(x-5)\log\, 4 = (2x-1) log\, 7$ and because it is linear equation you can proceed to finish it
 
Please, show details how you got (x−5)log4=(2x−1)log7
 
Nikolas7 said:
Please, show details how you got (x−5)log4=(2x−1)log7

this is as per definition $\log\,a ^ x = x \log\, a $
 
Yes, I figured out, thanks
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Replies
7
Views
1K
Replies
1
Views
1K
Replies
6
Views
1K
Replies
4
Views
1K
Replies
8
Views
1K
Replies
2
Views
2K
Replies
5
Views
2K
Replies
14
Views
2K
Replies
3
Views
1K
Back
Top