Answer:
- domain: all terms are defined for <em>all real numbers</em>
- solution: x = 6
Step-by-step explanation:
Rewrite the equation as a single exponential. After taking the log, the solution becomes obvious.
![5^{x-2}-7^{x-3}=7^{x-5}+11\cdot 5^{x-4}\\\\5^{x-2}-11\cdot 5^{x-4}=7^{x-3}+7^{x-5}\\\\5^{x-2}(1-11\cdot 5^{-2})=7^{x-3}(1+7^{-2})\\\\5^{x-2}\dfrac{14}{25}=7^{x-3}\dfrac{50}{49}\\\\1=\dfrac{2\cdot 7^{x-3}5^4}{2\cdot 5^{x-2}7^3}=\left(\dfrac{7}{5}\right)^{x-6}\\\\0=(x-6)\log(1.4)\\\\x=6](https://tex.z-dn.net/?f=5%5E%7Bx-2%7D-7%5E%7Bx-3%7D%3D7%5E%7Bx-5%7D%2B11%5Ccdot%205%5E%7Bx-4%7D%5C%5C%5C%5C5%5E%7Bx-2%7D-11%5Ccdot%205%5E%7Bx-4%7D%3D7%5E%7Bx-3%7D%2B7%5E%7Bx-5%7D%5C%5C%5C%5C5%5E%7Bx-2%7D%281-11%5Ccdot%205%5E%7B-2%7D%29%3D7%5E%7Bx-3%7D%281%2B7%5E%7B-2%7D%29%5C%5C%5C%5C5%5E%7Bx-2%7D%5Cdfrac%7B14%7D%7B25%7D%3D7%5E%7Bx-3%7D%5Cdfrac%7B50%7D%7B49%7D%5C%5C%5C%5C1%3D%5Cdfrac%7B2%5Ccdot%207%5E%7Bx-3%7D5%5E4%7D%7B2%5Ccdot%205%5E%7Bx-2%7D7%5E3%7D%3D%5Cleft%28%5Cdfrac%7B7%7D%7B5%7D%5Cright%29%5E%7Bx-6%7D%5C%5C%5C%5C0%3D%28x-6%29%5Clog%281.4%29%5C%5C%5C%5Cx%3D6)
The askser is 283828288288 god it’s da key bro
Answer:
y = (5/4)2^x
Step-by-step explanation:
The function value increases by a factor of 40/10 = 4 when x increases by 2. The function can be written as ...
y = (reference value)·(growth factor)^((x -reference)/(change in x for growth factor))
y = 10·4^((x-3)/2) . . . . . . using point (3, 10) as a reference
This can be simplified to ...
y = 10·2^(x -3) = 10/8·2^x
y = (5/4)2^x
The question doesn’t make much sense. Do you mean x=50-12? If so, x=38
If .65 were to repeat it would be .656565656565656