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

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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.
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Help me with the following equation:
${4}^{x-5}$=${7}^{2x-1}$
 
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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
 

Thread 'Trigonometry problem of interest'

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...
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