How do you solve #8^x=1000#?

2 Answers
AA_123 · EZ as pi
Nov 28, 2016

#x~~3.322#

Explanation:

Convert the exponential form to a logarithmic form #a^x=b-> x=log_ab#

#8^x=1000#

#x=log_8 1000#

You can use the 'change of base law' to calculate it.

#log_a b = (log_c b)/(log_c a)" "# (c is usually 10)

#x = log_10 1000//log_10 8#

#x~~3.322#

Jim G.
Nov 29, 2016

#x≈3.322#

Explanation:

Use the #color(blue)"law of logarithms"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(logx^n=nlogx)color(white)(2/2)|)))#
Applies to logarithms to any base.

Take the ln ( natural log) of both sides.

#rArrln8^x=ln1000#

Using the above law.

#rArrxln8=ln1000#

divide both sides by ln8

#(x cancel(ln8))/cancel(ln8)=ln1000/ln8#

#rArrx≈3.322" to 3 decimal places"#

Related questions
See all questions in Logarithmic Models
Impact of this question
6813 views around the world
You can reuse this answer
Creative Commons License