Solve Algebra 2 Problem 43: 7^3x/2^7x-1=(43.2)^x+2

[mustang]

Problem 43.
Solve and check.
7^3x divided by 2^7x-1 = (43.2)^x+2

 

[HallsofIvy]

I am assuming this is that you meant
\frac{7^{3x}}{2^{7x-1}}= 43.2^{x+2}
rather than
\frac{7^3x}{2^7x-1}= 43.2^x+ 2
which would be harder.

The key step is to use the logarithm to get rid of the "exponentials" (It really doesn't matter which base logarithm you use):
log(\frac{7^{3x}}{2^{7x-1}})= log(43.2^{x+2})
which is:
log(7^{3x})- log(2^{7x-1})= log(34.2^{x+2})
3x log(7)- (7x-1)log(2)= (x+2)log(34.2)<br /> <br /> That&#039;s now a simple linear equation with some peculiar coefficients.

 

[M98Ranger]

To solve this problem, we will first use the properties of exponents to rewrite the equation in a simpler form. We know that (a^m)^n = a^(m*n) and (a/b)^n = a^n/b^n. Using these properties, we can rewrite the equation as:

(7^3)^x / (2^7)^x-1 = (43.2)^x+2

Now, we can simplify further by evaluating the exponents on the left side of the equation:

7^(3x) / 2^(7x-1) = (43.2)^x+2

Next, we can use the power rule of exponents to rewrite 43.2 as (432/10) and raise both sides of the equation to the power of x+2:

7^(3x) / 2^(7x-1) = (432/10)^(x+2)

Using the property of exponents (a^m)^n = a^(m*n), we can rewrite the right side of the equation as:

7^(3x) / 2^(7x-1) = (432/10)^x * (432/10)^2

Now, we can simplify the right side of the equation by evaluating (432/10)^2 as 432^2/10^2:

7^(3x) / 2^(7x-1) = (432/10)^x * (432^2/10^2)

Next, we can use the power rule of exponents again to rewrite the right side of the equation as:

7^(3x) / 2^(7x-1) = (432/10)^x * 432^(2x)/10^(2x)

Finally, we can use the property of exponents (a/b)^n = a^n/b^n to rewrite the right side of the equation as:

7^(3x) / 2^(7x-1) = (432/10 * 432^(2x)) / (10 * 10^(2x))

Now, we can simplify the right side of the equation by combining like terms:

7^(3x) / 2^(7x-1) = (432 * 432^(2x)) / (10 * 10^(2x+1))

To solve for x, we can take the natural logarithm of both sides of