Thread: Common logarithims?

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

Originally Posted by homeylova223

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 ... parenthesis, remember?

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)

3x log(.16) + x log(.3) = 8 log(.3) - 4 log(.16)

factor out an x on the left side and solve.

.

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

Never mind I understand now