Logarithm

[sweetcharms]

how do u go about the PROOF that a^log c base b = a^log a base b?

[Math Tutor]

I am not shure I understand you.

You want to prove that:

[tex]a^{log_b c} = a^{log_ba}[/tex]

That is not true.
It is true only if c=a

[dduclam]

I am not shure I understand you.

You want to prove that:

[tex]a^{log_b c} = a^{log_ba}[/tex]

That is not true.
It is true only if c=a

This is true: [tex]a^{log_b c} = c^{log_ba}[/tex]

[charlie_eppes]

[tex]2^{log_{2}(x^{2}+7)}=6x+2[/tex]
Little help?

[broniran]

[tex]2^{log_{2}(x^{2}+7)}=6x+2[/tex]
Little help?

Use [tex]a^{log_a b}=b[/tex]

[Guest]

[tex]2^{log_{2}(x^{2}+7)}=6x+2[/tex]
Little help?

Use [tex]a^{log_a b}=b[/tex]

X^(2)+7 = 6x+2 ( a^log a (b)= b )
X^(2)=6×+2-7
X^(2)=6x-5
X^(2)-6x+5=0
X^(2)-×-5×+5=0
X(x-1)-5(x-1)=0
(X-1)(x-5)=0
Therefore x=1 (or) x=5

[Guest]

Or
[tex]x^2- 6x+ 5= 0[/tex]
[tex](x- 5)(x- 1)= 0[/tex]
Either x- 5= 0 so x= 5 or x- 1= 0 so x= 1.

[Guest]

To prove that a^log c base b=c^log a base b
we can use the change of base formula for logarithms and some properties of exponents.
Given:
a^log c base b

We know that
log c base b = log c base a/log b base a (Change of base formula).
So, we can rewrite the expression as:
a(log c base a/ log b base a)
Using the property of exponents
(a^m)^n=a^mn, we can rewrite the expression as:
(a^ log c base a)1/log b base a
(c)^ 1/log b base a
Now, we need to find c^log b base a:
c^log b base a =(b^log c base b)^log a base b
(b^ log b base a)^ log c base b
(a)^log c base b
Now, comparing this result with the expression we got earlier, we see that they are equal:
c^ log a base b =(a)^ log c base b
Hence, a^log c base b=c^log a base b is proved.

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