1) log(9 – x) – log(6 + x) = 1
2) e^2x + e^2x = 1
3) 2e^6x + e^6x = b
log [(9-x)/(6+x) ] = 1
if that is log base ten then
(9-x)/(6+x) = 10^1 = 10
9-x = 60 + 10 x
solve for x
e^(4x^2) = 1
ln (e^4x^2) = ln 1
4 x^2 = 0
x = 0
3 e^6x = b
e^6x = (b/3)
6x = ln b - ln 3
x = (1/6 ) (ln b/3)
1) To solve the equation log(9 – x) – log(6 + x) = 1, we can use the properties of logarithms to simplify it.
First, we can rewrite the equation using the quotient rule of logarithms:
log((9 - x) / (6 + x)) = 1
Next, we can convert the equation to exponential form by rewriting it as:
(9 - x) / (6 + x) = 10^1
Simplifying further, we have:
(9 - x) / (6 + x) = 10
Now, we can cross-multiply to eliminate the fractions:
9 - x = 10(6 + x)
Solve the equation by distributing 10:
9 - x = 60 + 10x
Combine like terms:
9 + 60 = 10x + x
69 = 11x
Divide both sides by 11:
x = 69/11
Therefore, the solution to the equation is x = 6.
2) The equation e^2x + e^2x = 1 can be addressed by combining the terms on the left side:
2e^2x = 1
Now, we can divide both sides by 2:
e^2x = 1/2
To solve for x, we can take the natural logarithm (ln) of both sides:
ln(e^2x) = ln(1/2)
By the properties of logarithms, we can simplify:
2x ln(e) = ln(1/2)
Since ln(e) is equal to 1:
2x = ln(1/2)
Finally, divide both sides by 2:
x = ln(1/2) / 2
Hence, the solution to the equation is x = ln(1/2) / 2.
3) To solve the equation 2e^6x + e^6x = b, first, combine the terms on the left side:
3e^6x = b
Next, divide both sides by 3:
e^6x = b/3
To solve for x, we can take the natural logarithm (ln) of both sides:
ln(e^6x) = ln(b/3)
Using the properties of logarithms, we get:
6x ln(e) = ln(b/3)
Since ln(e) is equal to 1:
6x = ln(b/3)
Lastly, divide both sides by 6:
x = ln(b/3) / 6
Thus, the solution to the equation is x = ln(b/3) / 6.