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Solve for $x.$$$4^{x}=23$$

$$2.26178$$

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 6

Properties of Logarithmic Functions

Campbell University

McMaster University

University of Michigan - Ann Arbor

Lectures

01:02

Solve: $x^{2}-4 x=3$

02:53

Solve for $x.$$$4^{x+2…

02:05

Solve.$x^{4}+3=4 x^{2}…

01:01

Solve for $x.$$$4^{3}=…

01:07

Solve.$$x^{2}+4 x=3$$<…

eso We're gonna use logs to solve this problem Forward to the X power equals 23 and ah ah, good Stone. Be able to guess pretty well that you know, four square to 16. So that's not quite good enough before cube to 64 s. Oh, this is just a thought bubble than my answer. Needs to be bigger than two, but smaller than three. Well, what we can do to solve this is actually take the log of both sides and you could do log or natural log. It doesn't actually matter. But the reason why this is the method to do the problem is because now we can use the power rule with the law of logs to move that X in front. Start looking at X Times log of four is equal to log of 23. Well, this is just a multiplication problem. So we solve multiplication problems with division and to get the approximate answer would be to divide. And that's where the calculator comes into play. Well, I've been 23. Divide by log four and you get about is approximate Healy here, 2.262 I liked around three decimals. It's 2.261780978 on my calculator. But this is what I'm going to write down, okay? It's

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