I suspect a typo, but ...
125x = (1/5)2x-3
625x = 2x - 15
623x = -15
x = -15/623
However, I suspect you may have meant
125^x = (1/5)^(2x-3)
5^(3x) = 5^(3-2x)
3x = 3-2x
5x = 3
x = 5/3
6x = 32
divide by 6
6^x = 32
x = log32/log6
log_x(10) = 1/2
10 = x^(1/2)
100 = x
If I interpreted any of these wrong, maybe my steps will allow you to solve them. Post your work if you get stuck.
) Solve for the values of x in the following equations. ?
i. 125x = (1/5)2x-3
ii. 6x = 32
iii. logx 10 = 1/2
2 answers
oops ...
5x = 3
x = 3/5
5x = 3
x = 3/5
1 answer