please show the way of solving in details.
We can use the rule that logA + logB = logAB.
Now, in the problem, this means that log8x²=2, or that 10² = 8x².
8x²=100
x²=25/2
x=5/√2=(5√2)/2
Hope this helps!
-MATH AVENGER
log[base 2](x) + log[base 4](x) = 2
First, we cannot combine logs until the logs are the same base. In order to make this the same base, use the change of base formula to change log[base 4] into log[base 2]. A reminder that
log[base c](a) = log[base b](a) / log[base b](c)
log[base 2](x) + log[base 2](x) / log[base 2](4) = 2
But log[base 2](4) = 2, so
log[base 2](x) + log[base 2](x) / 2 = 2
Multiply everything by 2, to get rid of the fraction.
2 log[base 2](x) + log[base 2](x) = 4
Move the 2 inside of the log, as per the log identity
c log[base b](a) = log[base b](a^c)
log[base 2](x^2) + log[base 2](x) = 4
Combine the logs.
log[base 2](x^2 x) = 4
log[base 2](x^3) = 4
2^4 = x^3
16 = x^3
Therefore,
x = (16)^(1/3)
x = (8 * 2)^(1/3)
X = 8^(1/3) 2^(1/3)
X = 2 * 2^(1/3)
X = 2^(4/3)
you can write the sum of two logs as the multiplication of the 2x and the 4x.
taking the antilog of both sides
(2x)*(4x) = antilog of 2
8x^2 = 100
x = 3.5355
Do you mean this:
log(2x) + log(4x) = 2
Then it is the same as:
(2x)(4x) = 100
x^2 = 12.5
x = 3.535533906
Check:
log(2x) + log(4x) = 0.849485002 + 1.150514998
... = 2
minorcho.... is correct but....he assumed base to be 10
I think value of base should be specified !!!1
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Verified answer
We can use the rule that logA + logB = logAB.
Now, in the problem, this means that log8x²=2, or that 10² = 8x².
8x²=100
x²=25/2
x=5/√2=(5√2)/2
Hope this helps!
-MATH AVENGER
log[base 2](x) + log[base 4](x) = 2
First, we cannot combine logs until the logs are the same base. In order to make this the same base, use the change of base formula to change log[base 4] into log[base 2]. A reminder that
log[base c](a) = log[base b](a) / log[base b](c)
log[base 2](x) + log[base 2](x) / log[base 2](4) = 2
But log[base 2](4) = 2, so
log[base 2](x) + log[base 2](x) / 2 = 2
Multiply everything by 2, to get rid of the fraction.
2 log[base 2](x) + log[base 2](x) = 4
Move the 2 inside of the log, as per the log identity
c log[base b](a) = log[base b](a^c)
log[base 2](x^2) + log[base 2](x) = 4
Combine the logs.
log[base 2](x^2 x) = 4
log[base 2](x^3) = 4
2^4 = x^3
16 = x^3
Therefore,
x = (16)^(1/3)
x = (8 * 2)^(1/3)
X = 8^(1/3) 2^(1/3)
X = 2 * 2^(1/3)
X = 2^(4/3)
you can write the sum of two logs as the multiplication of the 2x and the 4x.
taking the antilog of both sides
(2x)*(4x) = antilog of 2
8x^2 = 100
x = 3.5355
Do you mean this:
log(2x) + log(4x) = 2
Then it is the same as:
(2x)(4x) = 100
8x^2 = 100
x^2 = 12.5
x = 3.535533906
Check:
x^2 = 12.5
log(2x) + log(4x) = 0.849485002 + 1.150514998
... = 2
minorcho.... is correct but....he assumed base to be 10
I think value of base should be specified !!!1