How do you solve #5 (2^(3x)) = 8#?

3 Answers
May 7, 2018

#x = \frac{log_2(\frac{8}{5})}{3}#

Explanation:

We must isolate the #x#: first of all, get rid of the #5#, dividing both sides by #5#:

#2^(3x) = \frac{8}{5}#

Now, take logarithm (base #2#) to both sides:

#3x = log_2(\frac{8}{5})#

Finally, divide both sides by #3#:

#x = \frac{log_2(\frac{8}{5})}{3}#

If you prefer, you can use the rule #log(a/b) = log(a)-log(b)# to write

#log_2(\frac{8}{5}) = log_2(8) - log_2(5)=3-log_2(5)#

And the answer becomes

#x = 1-\frac{log_2(5)}{3}#

May 7, 2018

#x = (log 1.6)/(3log(2)) = (ln 1.6)/(3ln(2)) ~~ .226#

Explanation:

Given: #5(2^(3x)) = 8#

First divide by #5: " "2^(3x) = 8/5 = 1.6#

Log base 2 both sides: #" " log_2 2^(3x) = log_2 1.6#

Use the logarithmic property: #log_b b^x = x#

#3x = log_2 1.6#

#x =( log_2 1.6)/3 = 1/3 ( log_2 1.6)#

Use the change of base formula to convert to either log base 10 or the natural log: #log_b x =( log x)/(log 2) = ( ln x)/(ln 2) #

#x = 1/3 (log 1.6)/(log(2)) = (log 1.6)/(3log(2)) ~~.226#

May 7, 2018

#x=1-1/3log_2(5)#

Explanation:

You need to work with exponents here.

#5*2^(3x)=8=2^3#
That means #2^(3x)=2^3/5#
Divide with #2^3# on both sides: #2^(3(x-1))=1/5#

Take #log_2# on both sides:

#3(x-1)=-log_2(5)#

or #x=1-1/3log_2(5)#

Check:
#5*2^(3(1-1/3log_2(5))# =#5*2^(3-log_2(5)#=#5*8/5=8#