To solve 49(3x) = 343(2x+1), write each side of the equation in terms of base what? .
In the brackets are powers.
I tried to solve this but I don't understand I believe it is the first base you go off from but I'm not sure what I'm doing and a reassurance and an explanation would be greatly appreciated.It is solving bases by rewriting bases
The answer is 7
7 is the answer
49^3x = 343^(2x+1).
3x*Log49 = (2x+1)*Log343,
Divide both sides by the Log of 49:
3x = (2x+1)*Log343/Log49,
3x = 1.5(2x+1), 3x = 3x+1.5,
3x-3x = 1.5. No solution?
Please make sure the problem is
copied correctly.
To solve the equation 49(3x) = 343(2x+1) in terms of a common base, we need to determine the base that both sides of the equation can be expressed in. In this case, the given numbers are 49 and 343, which are both powers of the same base.
To identify the base, we find the common factor between the two numbers:
49 = 7^2
343 = 7^3
Here, we see that both numbers can be expressed as powers of 7. Therefore, we can rewrite the equation in terms of base 7:
7^2(3x) = 7^3(2x+1)
Next, we can simplify the equation by applying the power of a product rule, which states that (a*b)^c = a^c * b^c. Applying this rule, we can rewrite the equation as:
7^(2+1)x = 7^(3)(2x+1)
Simplifying further, we have:
7^3x = 7^3(2x+1)
Now, it is clear that the base for both sides of the equation is 7.
I got x=1/2939
But that was incorrect.
Now I believe it's just one
49^3x = 343^(x+1),
3x*Log49 = (x+1)*Log343,
Divide both sides by Log49:
3x = (x+1)*Log343/Log49,
3x = 1.5*(x+1),
3x = 1.5x + 1.5,
3x-1.5x = 1.5,
X = 1.
So the 2x in your Eq should be x.