Solving logarithmic equations help! Easy problem!

Can someone help me solve this problem? Like show the steps :)) thank you!

I got the answer: 4 - log5/log3

3^4-x = 5

I'm going to assume you mean 3 ^ (4-x) = 5, otherwise it wouldn't be a log question!

Take the log of both sides:
log (3^(4-x)) = log 5

Bring the power to the front of the log
(4-x) log 3 = log 5

Divide both sides by log 3
4 - x = log 5 / log 3

Subtract 4 from both sides
-x = log 5 / log 3 - 4

Multiply both sides by -1
x = 4 - log 5 / log 3

x = 2.535...

So you're right!

It's important to specify that the exponent is 4 - x this way:

3^(4-x) = 5

Therefore:

log [3^(4-x)] = log 5

(I'm using base 10)

(4 - x)log 3 = log 5

(rule of logarithms with exponents)

4 - x = log 5 / log 3

x = 4 - (log 5 / log 3)

I hope this helps!

I'm assuming the 4-x is the entire exponent, as in 3^(4 - x) = 5

(4 - x) log 3 = log 5
(4 - x) = (log 5)/(log 3)
-x = (log 5)/(log 3) - 4
x = 4 - (log 5)/(log 3)

Your answer is correct.

Hi,

3^(4-x) = 5

log 3^(4-x) = log 5

(4 - x)log 3 = log 5

(4 - x)log 3 = log 5

4 log 3 - x log 3 = log 5

4 log 3 - log 5 = x log 3

4 log 3 - log 5
-------------------- = x <==ANSWER
. . . log 3

4 - log 5/log 3 = x <==ANSWER

Your answer is correct!! :-) I hope that helps!! :-)