At 2:54 you introduce 125-5 You have effectively guessed at m=5 without giving a clue as to how you got that number. What if the rhs was 60. Then we could guess 64-4. But what about 100. but if we guess m=5 as a solution. then no need for 4 minutes to prove it. all we need to show is that m=5 is unique. m^3-m = m(m^2-1) m is positive increasing for m>1, m^2-1 is positive and increasing. thus for m>1 lhs is positive increasing and rhs is constant so there is only one real solution which is m=5 so 2^logx=5 (1) logx.log2=log5 logx=log5/log2 (2) but 5.2 =10 log5 + log2 =log10=1 log5=1-log2 subs this into eqn(2) logx=( 1-log2)/log2 logx = 1/log2 -1 raise to power 10 x= 10^(1/log2) /10 = 2098.59239587 /10 x= 209.859239587 no need for other base logs
8^logx-2^logx=5! 2^logx(2^2logx-1)=5.4.3.2.1 2^logx(2^logx+1)(2^logx-1)=6.5.4 By comparison 2^logx=5 Take log of both sides Logx=log5/log2=2.32192 X=antilog2.32192
^=read as to the power *=read as square root As per question 8^(logx)- 2^(logx)=5! {(2^3)^logx)}- 2^(logx)=120 2^(3logx) - 2^(logx)=120 Let 2^(logx)=R So, R^3-R=(5^3)-5 So, R=5 2^(logx)=5 Take the log log2^(logx)=log5 logx. log2=log5 logx= log5/log2 X= antilog { log5/log2}...May be
Respected Sir, Good evenings. Nicely solved...
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Let x=2^y. Then (2^y)^3 -2^y-120=0. Clearly 2^y=5 and y=log(5)/log(2). Then x=e^(log (5)/log (2))=10.1953
At 2:54 you introduce 125-5
You have effectively guessed at m=5 without giving a clue as to how you got that number.
What if the rhs was 60. Then we could guess 64-4. But what about 100.
but if we guess m=5 as a solution. then no need for 4 minutes to prove it.
all we need to show is that m=5 is unique.
m^3-m = m(m^2-1)
m is positive increasing
for m>1, m^2-1 is positive and increasing.
thus for m>1 lhs is positive increasing and rhs is constant so there is only one real solution which is m=5
so
2^logx=5 (1)
logx.log2=log5
logx=log5/log2 (2)
but 5.2 =10
log5 + log2 =log10=1
log5=1-log2 subs this into eqn(2)
logx=( 1-log2)/log2
logx = 1/log2 -1
raise to power 10
x= 10^(1/log2) /10
= 2098.59239587 /10
x= 209.859239587
no need for other base logs
8^logx-2^logx=5!
2^logx(2^2logx-1)=5.4.3.2.1
2^logx(2^logx+1)(2^logx-1)=6.5.4
By comparison 2^logx=5
Take log of both sides
Logx=log5/log2=2.32192
X=antilog2.32192
^=read as to the power
*=read as square root
As per question
8^(logx)- 2^(logx)=5!
{(2^3)^logx)}- 2^(logx)=120
2^(3logx) - 2^(logx)=120
Let 2^(logx)=R
So,
R^3-R=(5^3)-5
So,
R=5
2^(logx)=5
Take the log
log2^(logx)=log5
logx. log2=log5
logx= log5/log2
X= antilog { log5/log2}...May be
8^[log(x)]-2^[log(x)]=5!
Domain: x≥1
Let n=log(x)
8ⁿ-2ⁿ=120
(2ⁿ)³-2ⁿ-120=0
Let h=2ⁿ
h³-h-120=0
h³-125-h+5=0
(h³-125)-(h-5)=0
(h-5)(h²+5h+25)-1(h-5)=0
(h-5)(h²+5h+24)=0
h²+5h+24=0
∆=b²-4ac
∆=5²-4•1•24
∆=25-96
∆=-71
I ❤ math
Решаем методом устного счета.
2^logx=y
y^3-y=120
y(у-1)(у+1)=120
120=5x4x6
y=5
2^log5=5
Good luck!
2^(log_10(x)=5 does not yield your result.
@davidseed2939 Напишите свое решение.
В Вашем решении
x=?
x= 210