solve for x: 3^2x+3=7^x+1
I did 2x+3=log0f3(7)+ logof3(7)
(2-log of 3(7))x=log of 3(7)-3
x=log of 3 (7)+log of 3 (3^-3)/log 0f 3 (3^2)-log of 3 (7)
x=log of 3 (7-1/3^3)/log of 3(9/7)
log of 3(7/27)/log of 3(9/7)
answer log of 9/7(7/27)
Is this correct? thank you for checking my work.
I assume you meant
3^(2x+3) = 7^(x+1)
Since the bases are in no way related, just take logs, in whatever base you like
(2x+3) log3 = (x+1) log7
2x log3 + 3log3 = x log7 + log7
(2log3 - log7)x = log7 - 3log3
x = (log7 - 3log3)/(2log3 - log7)
or, if you don't mind fractions,
x = log(7/27) / log(9/7)
so, you were right, though complicated, until your final step. In the expression "log x" the "log" part is not a factor that can cancel. If you want a "simpler" form, remember the base change formula, and you actually have
log9/7 (7/27)
Thank you
Let's check your work step by step:
Starting with the equation: 3^(2x+3) = 7^(x+1)
First, we can simplify the equation by using logarithms:
1. Take the logarithm of both sides of the equation using the same base. It seems you chose base 3 logarithm.
log3(3^(2x+3)) = log3(7^(x+1))
2. Apply the exponent rule to move the exponent down in front:
(2x+3)log3(3) = (x+1)log3(7)
We know that log3(3) = 1, so we can proceed:
3. Simplify the expression:
2x + 3 = (x + 1)log3(7)
It seems there is a mistake in your calculations from this point forward.
To continue, we move all terms with x to one side and the constants to the other side:
4. Move the x term on one side and the constant term on the other side:
2x - x = (x + 1)log3(7) - 3
The equation becomes:
x = (x + 1)log3(7) - 3
Now, let's solve for x. We'll distribute log3(7) to both terms inside the parentheses:
x = xlog3(7) + log3(7) - 3
We gather like terms:
x - xlog3(7) = log3(7) - 3
Now, we can factor out x:
x(1 - log3(7)) = log3(7) - 3
Finally, divide by (1 - log3(7)) to solve for x:
x = (log3(7) - 3) / (1 - log3(7))
This is the simplified expression for x in terms of logarithms.
I suggest rechecking your calculations to ensure accuracy and correctness.