Let a & b be real numbers, with a & b > 1 solve... LOGS Help

[dhuff]

Let a & b be real numbers, with a & b > 1 solve
log_b(x^a) = (log_bx)<sup>a

Can anyone help me solve this? I need step by step instructions!

Thanks</sup>

 

[Deleted member 4993]

Let a & b be real numbers, with a & b > 1 solve
log_b(x^a) = (log_bx)<sup>a

Can anyone help me solve this? I need step by step instructions!

Thanks</sup>

Hint:

log_b(x^a) = a *(log_bx)


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[soroban]

Hello, dhuff!

\(\displaystyle a\text{ and }b\text{ are real numbers, with }a,b > 1.\)

\(\displaystyle \text{Solve: }\:\log_b(x^a) \:=\:\left(\log_bx\right)^a\)

We have: .\(\displaystyle \left(\log_bx\right)^a - \log_b(x^a) \:=\:0 \)

. . . . . . . \(\displaystyle (\log_bx)^a - a(\log_bx) \:=\:0\)

. . . . . \(\displaystyle \log_bx\left[(\log_bx)^{a-1} - a\right] \:=\:0\)


And we have two equations to solve:

\(\displaystyle \log_bx \:=\:0 \quad\Rightarrow\quad x \:=\:b^0 \quad\Rightarrow\quad \color{blue}{x \:=\:1}\)

\(\displaystyle (\log_bx)^{a-1} - a \:=\:0 \quad\Rightarrow\quad (\log_bx)^{a-1} \:=\:a \)
. . . . \(\displaystyle \log_bx \:=\:a^{\frac{1}{a-1}} \quad\Rightarrow\quad \color{blue}{x \:=\:b^{a^{^{\frac{1}{a-1}}}}} \)