If log\(_810\) = X, evaluate log\(_85\) in terms of X.
\(\frac{1}{2}\)X
X-\(\frac{1}{4}\)
X-\(\frac{1}{3}\)
X-\(\frac{1}{2}\)
Explanation
Video Explanation
No video available
Post your Contribution
Discussions (196)
Hmm,my people,let's just think. If we simplify log 5 base8 log 2 base 8,we will get log 10 base 8 and we have been given that log 10 base 8 = X. Now let's move on.
Log 5 base 8 log 2 base 8 = X. Are u with me?,,we must be careful here. Can you still remember the rule that says loga^2 base a^3 gives 2/3 log a base a which is eaqual to 2/3 X 1 = 2/3.
Can you still undstd. Go along with me my brother and sisster,
so let's apply d same thing here, log 2 ^1 base 8. Do not forget that 8 can also be written as 2^3.
You got it? If u are with me,am sure u're undstding a lilltle bit.
So we have log 2^1 base 2 ^3 to be 1/3 log 2 base 2 which will now give us 1/ 3 X 1 = 1/3.
U understand me,don't relax pls,keep going along with me.
Recall that we say log5 base 8 X log 2 base 8 = X.
Now that we know our log 2 base 8 = 1/3,so we put into d equation to have..
Log 5 base 8 1/ 3 = X.
Therefore,log 5 base 8 = X - 1/3.
Log10 to base8=X is given.
Log5 to base 8= log(10/2) to base 8.
Then log10-log2= X-log2. (-log 2 to base 8=Y.) 8(power of y)=2
2 (power of 3y)=2(power of 1).
Eliminate 2, left with 3y=1.
Y=1/3. Then bringing down the - negative sign. The answer is Y=-(1/3).
If log10 to base 8 = X
I.e log5 - base8
Log(5 - 8)
X = log3
Applying the law of indices which if 3^0 = 1, a = 1
So log3 = 1
Applying the rule of zero power law
Log3^-1 = 1/3 I.e X=log1/3
I think it can also be solved this way:
Log 10 base8=X...(1).Equation(1) can also be written as: log 2 base8 + log5 base8=X...(1)....1st law of log. .
Transfering log2 base8 to the R.H.S We have: log5 base8=X- Log2 base8....(*)
Solving for log2 base8 in (*) above, we have 1/3. Therefore substituting 1/3 into equation (*), we have:
log 5 base8=X - [1/3]
since log5 to base 8 = x, and x = log10 to base 8
i.e log5 to base 8 =log10 to base8
then x =log5 to base 8 - log10 to base 8, 5/10=1/2,i.e 2^-1 and 8=2^3,= -1/3=x
Woahhhhhh ! All cool..............
Log 10 base 8 = X
log 5 base 8 = log 10/2
" law of log
log 10 base 8 - log 2 base 8
since log 10 is 'X' lets equate log 2 base 8 to 'Y'
log 2 base 8 = Y
2^1 = 8^y ==== 2^1 = 2^3y
equating the powers
1=3y ===== y = 1/3
recall X - Y
log 5 base 8 = X-(1/3)
it can also be done this way
log10 to base 8 is the same as log5x2 base 8
therefore log5+log2=x,
so, log2 to base 8 is the same as log 2 to base 2 raise to the power of 3
then, 1/3log2 to base 2 will be 1/3
since lo5 to base 3 is x, then x+1/3 will be x=-1/3.
To solve the question, follow the steps:
1. Log10 to base 8 is equal to x.
2. 10 is same as 2×5, so in that case, rewrite log10 to base 8 as log(2×5) to base 8.
3. According to the laws of logarithm, rewrite log(2×5) to base 8 as log2 to base 8 + log5 to base 8. This is the main equation.
4. Equate log2 to base 8 as y.
5. Convert the logarithm to indices i.e. log2 to base 8 as y becomes 2^1=8^y
6. 8 is same as 2^3:. 2^1=8^y becomes 2^1=2^(3y).
7. Equate the powers by cancelling out the bases. So it remains 1=3y.
8. Divide both sides by 3. Then y=1/3.
9. Substitute the value of y in the main equation i.e. (1/3)+log5 to base 8 = x.
10. Take 1/3 to the other side i.e. log5 to base 8 = x-(1/3). That is option C.
Log8 raise to power 10 minus log8 raise to power 5 equal to X gives log8 raise to power 2 equal to X. Therefore, 8 raise to power x = 2 raise to power one. Then, 2 raise to power 3x= 2 raise to power one . Then, 2 cancel on both sides wht will b left is 3x=1. X=1\3 when X crosses equality sign ,d ansa will b X-1\3. Tanx nd God bless una
Log8 10=X is given.
Log8 5 = log8(10/2).
Then log10 - log2 = X - log2.
(-log8 2 =X) 8(x) =2
2^(3y) = 2^(1)
Eliminate 2, e.g (2^((3x)) =2^(1)) Then you are left with 3x =1.
= 3x =1
x=1/3
.∴ x=-1/3
By bringing down the - negative sign. The answer is x=-(1/3)
