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I am given the expression $3^{4x-5}=38$ and asked to rewrite in common log to isolate the variable $x$, then to solve using a calculator.

I am struggling to to get the $x$ on it's own. My attempt:

$$3^{4x-5}=38$$

Rewrite lhs as a log

$$log_3(4x-5)=38$$

Rewrite lhs into common log base 10:

$$\frac{log(4x-5)}{log(3)}=38$$

This is as far as I can go. How can I isolate x here?

[Edit]

I rewrote to this, does it look right?

$$3^{4x-5}=38$$ $$log_3(38)=4x-4$$

My textbook says I am specifically to rewrite using the common log: $$\frac{log(38)}{log(3)}+5=4x$$ $$x=\frac{(\frac{log(38)}{log(3)}+5))}{4}$$

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    $\begingroup$ $a^b=c$ gives $b\ln(a)=\ln(c)$ so there is a mistake both on LHS and RHS. Equivalently you could have written $\log_3(38)=4x-5$. $\endgroup$
    – zwim
    Jan 23 at 14:42
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    $\begingroup$ After your edit, this is correct. $\endgroup$
    – zwim
    Jan 23 at 15:59
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Hint: $\log a^b = b \log a$ and not $\log_ba$

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