How do you solve #5^x=52#?

3 Answers
Jun 1, 2018

Answer:

#x=log_(5)52#

Explanation:

#5^x=52#

Taking log both sides with base #5#,

#x=log_(5)52#

Jun 1, 2018

Answer:

# x = (log (52 ))/log(5) = 2.4550457 ...#

Explanation:

We seek a solution of:

# 5^x = 52 #

We first note that there is no trivial solution, as such we must utilise logarithms to solve the equation. We can take logarithms of both sides (the base of the logarithm is arbitrary, as will be shown):

# log(5^x) = log (52 )#

And utilising a properties of logarithms, #loga^b=bloga# we have:

# x log(5) = log (52 )#

And re-arranging we get the solution:

# x = (log (52 ))/log(5)#

As indicated above, we have made no assumptions regarding the base of the logarithm, thus:

Using base #10# logarithms (in conjunction with a calculator):

# x = (log_10(52 ))/(log_10(5)) = (1.7160033 ...)/(0.6989700...) = 2.4550457 ...#

Using base #e# logarithms (Natural, or Napier logarithms):

# x = (ln(52 ))/(ln(5)) = (3.9512437 ...)/(1.6094379...) = 2.4550457 ...#

Using base #5# logarithms:

# x = (log_5(52 ))/(log_5(5)) = (2.4550457 ...)/(1) = 2.4550457 ...#

Jun 3, 2018

Answer:

#x~~2.46#

Explanation:

Whenever we have an exponential equation of the form

#a^x=b#

we can use the Change of Base Formula

#x=log(b)/log(a)#

In our situation, we essentially have #a=5# and #b=52#. Plugging into the formula, we get

#x=log52/log5#

#x~~2.46#

Hope this helps!