10^3x-8=2^5-x
I have a sneaking suspicion you might mean
10^(3x-8) = 2^(5-x)
???????????????
Yes
Been there, done that
http://www.jiskha.com/display.cgi?id=1452547715
log 10 [10^(3x-8) ] = 3x-8
log 10 [ 2^(5-x)] = (5-x)log 10[2]
so
3x - 8 = (5-x)(0.301)
To solve the equation 10^(3x-8) = 2^(5-x), we can start by using the property of logarithms that states: if a^b = c, then log_a(c) = b.
Let's apply this property to the given equation:
10^(3x-8) = 2^(5-x)
Taking the logarithm of both sides of the equation will allow us to solve for x:
log(10^(3x-8)) = log(2^(5-x))
Now we can simplify using the power rule of logarithms, which states: log_a(b^c) = c * log_a(b):
(3x-8) * log(10) = (5-x) * log(2)
The logarithm of 10 to any base is typically denoted as log(10) = 1, and the logarithm of 2 to any base is typically denoted as log(2) = 0.301.
Therefore, the equation becomes:
(3x-8) = (5-x) * 0.301
Next, we can distribute the 0.301 to the terms on the right side of the equation:
3x - 8 = 0.301 * (5-x)
Simplifying further:
3x - 8 = 1.505 - 0.301x
Now we can solve for x by simplifying and rearranging the equation:
3x + 0.301x = 1.505 + 8
Combining like terms:
3.301x = 9.505
Dividing both sides of the equation by 3.301:
x = 9.505 / 3.301
x ≈ 2.879
Therefore, the solution to the equation 10^(3x-8) = 2^(5-x) is approximately x = 2.879.