4^x-2=3 How do I solve this and round my answer 3 places past the decimal?
回答 (9)
2021-04-14 1:01 am
Case 1:
4^(x-2)=3
=>
(x-2)ln(4)=ln(3)
=>
x=2+ln(3)/ln(4)
=>
x=2.792
Case 2:
4^x=5
=>
xln(4)=ln(5)
=>
x=ln(5)/ln(4)
=>
x=1.161
Mistake: if you meant (4^x)-2=3, then you had better writing
4^x=5; if not, you should write 4^(x-2)=3. Inserting in a proper
bracket is a must.
4^(x-2)=3
=>
(x-2)ln(4)=ln(3)
=>
x=2+ln(3)/ln(4)
=>
x=2.792
Case 2:
4^x=5
=>
xln(4)=ln(5)
=>
x=ln(5)/ln(4)
=>
x=1.161
Mistake: if you meant (4^x)-2=3, then you had better writing
4^x=5; if not, you should write 4^(x-2)=3. Inserting in a proper
bracket is a must.
2021-04-13 2:29 am
4^x - 2 = 3
4^x = 3 + 2 = 5
xlog(4) = log(5)
x = log(5)/log(4)
Use a calculator:
x = 1.161
4^x = 3 + 2 = 5
xlog(4) = log(5)
x = log(5)/log(4)
Use a calculator:
x = 1.161
2021-04-13 2:25 am
As written:
4^x - 2 = 3
Add 2 to both sides:
4^x = 5
Base-4 log of both sides:
x = log₄(5)
base change so we can use a calculator:
x = ln(5) / ln(4)
x = 1.161 (rounded to 3DP)
--------
If you really meant:
4^(x - 2) = 3
Base-4 log, first:
x - 2 = log₄(3)
Add 2 to both sides:
x = log₄(3) + 2
Base change so we can use a calculator:
x = ln(3) / ln(4) + 2
x = 0.792 + 2 (quotient of logs rounded to 3DP)
x = 2.792
Proper use of parenthesis is important so everyone understands what you really meant for the question.
4^x - 2 = 3
Add 2 to both sides:
4^x = 5
Base-4 log of both sides:
x = log₄(5)
base change so we can use a calculator:
x = ln(5) / ln(4)
x = 1.161 (rounded to 3DP)
--------
If you really meant:
4^(x - 2) = 3
Base-4 log, first:
x - 2 = log₄(3)
Add 2 to both sides:
x = log₄(3) + 2
Base change so we can use a calculator:
x = ln(3) / ln(4) + 2
x = 0.792 + 2 (quotient of logs rounded to 3DP)
x = 2.792
Proper use of parenthesis is important so everyone understands what you really meant for the question.
2021-04-15 5:05 am
The basic premise of algebra is: if two things are equal, then any operation you do to both of them will produce new things that are still equal.
For example, if x = 3 then cos(log(x^17)) = cos(log(3^17))
Therefore your goal is to perform operations on both sides of the equation until one side is x alone.
4^x - 2 = 3
The obvious first step is to add 2 to both sides.
4^x - 2 + 2 = 3 + 2
4^x = 5
Now, what operation turns 4^x into x? The logarithm base 4.
log_4(4^x) = log_4(5)
x = log_4(5)
That's your answer! Next the question of how to type it into a calculator so as to get a rounded-off decimal. For that we use the fact that log_b(a) = log_c(a) / log_c(b). So type log(5)/log(4).
For example, if x = 3 then cos(log(x^17)) = cos(log(3^17))
Therefore your goal is to perform operations on both sides of the equation until one side is x alone.
4^x - 2 = 3
The obvious first step is to add 2 to both sides.
4^x - 2 + 2 = 3 + 2
4^x = 5
Now, what operation turns 4^x into x? The logarithm base 4.
log_4(4^x) = log_4(5)
x = log_4(5)
That's your answer! Next the question of how to type it into a calculator so as to get a rounded-off decimal. For that we use the fact that log_b(a) = log_c(a) / log_c(b). So type log(5)/log(4).
2021-04-14 4:53 pm
4^x=5
x=log(5)/log(4)=1.161
x=log(5)/log(4)=1.161
2021-04-13 5:51 am
4^(x) - 2 = 3 ?
4^(x) = 5
Ln[4^(x)] = Ln(5)
x.Ln(4) = Ln(5)
x = Ln(5)/Ln(4)
x ≈ 1.16096
x = 1.161
4^(x - 2) = 3 ?
Ln[4^(x - 2)] = Ln(3)
(x - 2).Ln(4) = Ln(3)
x - 2 = Ln(3)/Ln(4)
x = 2 + [Ln(3)/Ln(4)]
x = [2.Ln(4) + Ln(3)]/Ln(4)
x = [Ln(4²) + Ln(3)]/Ln(4)
x = [Ln(16) + Ln(3)]/Ln(4)
x = [Ln(16 * 3)]/Ln(4)
x = Ln(48)/Ln(4)
x ≈ 2.7924
x = 2.793
4^(x) = 5
Ln[4^(x)] = Ln(5)
x.Ln(4) = Ln(5)
x = Ln(5)/Ln(4)
x ≈ 1.16096
x = 1.161
4^(x - 2) = 3 ?
Ln[4^(x - 2)] = Ln(3)
(x - 2).Ln(4) = Ln(3)
x - 2 = Ln(3)/Ln(4)
x = 2 + [Ln(3)/Ln(4)]
x = [2.Ln(4) + Ln(3)]/Ln(4)
x = [Ln(4²) + Ln(3)]/Ln(4)
x = [Ln(16) + Ln(3)]/Ln(4)
x = [Ln(16 * 3)]/Ln(4)
x = Ln(48)/Ln(4)
x ≈ 2.7924
x = 2.793
2021-04-13 3:53 am
4^x - 2 = 3
4^x = 3 + 2
4^x = 5
x = (ln(5) + 2*i*pi*n) / ln(4), for any integer n
x =~ 1.161 + 4.532in, for any integer n
If x is a real number, then n = 0, so x = log[4](5) =~ 1.161
4^x = 3 + 2
4^x = 5
x = (ln(5) + 2*i*pi*n) / ln(4), for any integer n
x =~ 1.161 + 4.532in, for any integer n
If x is a real number, then n = 0, so x = log[4](5) =~ 1.161
2021-04-13 2:38 am
4^x - 2 = 3
4^x = 5
x = log(5)/log(4)
x ≈ 1.1610
4^x = 5
x = log(5)/log(4)
x ≈ 1.1610
2021-04-13 2:21 am
solution is $\text{ }\frac{\ln \left(5\right)}{2\ln \left(2\right)}$
收錄日期: 2021-04-24 08:40:30
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