Solve each equation or inequality
Can anyone offer assistance for the following two problems
1. 5^x=4^x+3 This is what I did
log 5^x=log 4^x+3
x log 5=x+3 log 4 Now I am stuck
2. .16^4x+3= .3^8-x
I am not entirely sure on how to start this one
Solve each equation or inequality
Can anyone offer assistance for the following two problems
1. 5^x=4^x+3 This is what I did
log 5^x=log 4^x+3
x log 5=x+3 log 4 Now I am stuck
2. .16^4x+3= .3^8-x
I am not entirely sure on how to start this one
Please fix your notation. You're all over the place on that on.
If you mean this: , you should have written 5^x = 4^(x+3). Parentheses make a difference.
This is okay, excepting the notation.
log [5^x] = log [4^(x+3)]
Again with notation confusion:
x log 5 = (x+3) log 4
Why are you stuck? Keep going!
x log 5 = x log(4) + 3 log 4
x log(5) - x log(4) = 3 log(4)
x [log(5) - log(4)] = 3 log(4)
Can you finish.
log(4) is a number. It may not be the prettiest number ever, but it's not the ugliest, either. Use it like any other number.
For my second problem I am having some difficulty (This is what I did)
.16^(4x+3)=.3^(8-x)
4+3x log .16= 8-x log .3
4 log .16 + 3x log .16= 8 log .3- x log .3
4 log .16+3x log .16+ x log .3= 8 log .3 (I am stuck right here) Then I did this
-3.18+(x)3 [ log .16 + log .3]= -4.183
-3.18- 3.956x=-4.183 But I think this is wrong somehow
I am still having some trouble solving it
3x log(.16)+ x log (.3)= 8 log (.3)- log(.16)
(x)3 log(.16)+ log(.3)=(4) log (.3)- log (.16) I multiplied 3 times log(.16) first
-2.9105x= -1.2956
x=.4451 But this is wrong because the answer in the back of my book says .3434
What could I be doing wrong