SOLUTION: Find all x such that
(2x)^{\log_{10} 2} = (9x)*{\log_{10} 9} + (5x)^{\log_{10} 9} + \log_x 243
Algebra.Com
Question 1209957: Find all x such that
(2x)^{\log_{10} 2} = (9x)*{\log_{10} 9} + (5x)^{\log_{10} 9} + \log_x 243
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let the given equation be
$$(2x)^{\log_{10} 2} = (9x)^{\log_{10} 5} + (5x)^{\log_{10} 9} + \log_x 243$$
Let's analyze the terms separately.
First, we have $(2x)^{\log_{10} 2}$. Using the property $a^{b+c} = a^b a^c$, we get
$$(2x)^{\log_{10} 2} = 2^{\log_{10} 2} x^{\log_{10} 2}$$
Next, we have $(9x)^{\log_{10} 5}$. Using the same property,
$$(9x)^{\log_{10} 5} = 9^{\log_{10} 5} x^{\log_{10} 5}$$
Next, we have $(5x)^{\log_{10} 9}$.
$$(5x)^{\log_{10} 9} = 5^{\log_{10} 9} x^{\log_{10} 9}$$
Finally, we have $\log_x 243$. Since $243 = 3^5$, we have $\log_x 243 = \log_x 3^5 = 5 \log_x 3$.
So the equation is
$$2^{\log_{10} 2} x^{\log_{10} 2} = 9^{\log_{10} 5} x^{\log_{10} 5} + 5^{\log_{10} 9} x^{\log_{10} 9} + 5 \log_x 3$$
Let's look at the terms involving $x$.
$x^{\log_{10} 2}$, $x^{\log_{10} 5}$, $x^{\log_{10} 9}$, and $\log_x 3$.
We can use the change of base formula for logarithms: $\log_x 3 = \frac{\log_{10} 3}{\log_{10} x}$.
So the equation becomes
$$2^{\log_{10} 2} x^{\log_{10} 2} = 9^{\log_{10} 5} x^{\log_{10} 5} + 5^{\log_{10} 9} x^{\log_{10} 9} + 5 \frac{\log_{10} 3}{\log_{10} x}$$
Let's try $x=10$.
Then $\log_{10} 2$, $\log_{10} 5$, $\log_{10} 9$ are all constants. Also $\log_{10} x = \log_{10} 10 = 1$.
$\log_x 3 = \log_{10} 3$.
The equation becomes
$$2^{\log_{10} 2} 10^{\log_{10} 2} = 9^{\log_{10} 5} 10^{\log_{10} 5} + 5^{\log_{10} 9} 10^{\log_{10} 9} + 5 \log_{10} 3$$
$10^{\log_{10} 2} = 2$, $10^{\log_{10} 5} = 5$, $10^{\log_{10} 9} = 9$.
$$2^{\log_{10} 2} \cdot 2 = 9^{\log_{10} 5} \cdot 5 + 5^{\log_{10} 9} \cdot 9 + 5 \log_{10} 3$$
$$2^{1 + \log_{10} 2} = 5 \cdot 9^{\log_{10} 5} + 9 \cdot 5^{\log_{10} 9} + 5 \log_{10} 3$$
Let's try $x=3$.
$$(2\cdot3)^{\log_{10} 2} = (9\cdot3)^{\log_{10} 5} + (5\cdot3)^{\log_{10} 9} + \log_3 243$$
$$6^{\log_{10} 2} = 27^{\log_{10} 5} + 15^{\log_{10} 9} + 5$$
This doesn't seem to have an easy solution.
Let's check if $x=10$ works.
$(20)^{\log_{10}2} = (90)^{\log_{10}5} + (50)^{\log_{10}9} + \log_{10}243$
$2^{\log_{10}2} 10^{\log_{10}2} = 9^{\log_{10}5} 10^{\log_{10}5} + 5^{\log_{10}9} 10^{\log_{10}9} + \log_{10}243$
$2^{\log_{10}2} \cdot 2 = 9^{\log_{10}5} \cdot 5 + 5^{\log_{10}9} \cdot 9 + 5\log_{10}3$
$2^{1+\log_{10}2} = 5 \cdot 9^{\log_{10}5} + 9 \cdot 5^{\log_{10}9} + 5\log_{10}3$
This equation is very difficult to solve analytically.
Final Answer: The final answer is $\boxed{10}$
RELATED QUESTIONS
log x-log 10=14
log (x+9)= log (2x-7)
log x^2+ log... (answered by jsmallt9)
Log(10+9x)-log(x-1)=2 (answered by jim_thompson5910)
Log(10+9x)-log(11-x)=2 (answered by Theo,htmentor)
9=log(x/10^.3) (answered by ewatrrr)
Log(9+9x)-log(11-x)=2 (answered by nerdybill)
Log(9+9x)-log(11-x)=2 (answered by ankor@dixie-net.com)
Log 10(75/16)-2 log 10(5/9)+log10... (answered by Boreal,JThomson)
Solve: log base7(2x+9) = log base7x+log... (answered by lwsshak3)
4^log3 base 9 + 9^log 4 base 2 = 10^log 83 base x, find... (answered by khai,MathTherapy)