I want to find the value of $x$ in this equation, by using logarithms.
$ \ 18^{4x-3} = (54 \sqrt{2})^{3x-4} \ $
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HINT:
$$18=(3\sqrt2)^2$$ and $$(3\sqrt2)^3=?$$
So, take logarithm in both sides with respect to base $3\sqrt2$
lab bhattacharjee
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It's 54√2 ! Got it! – Fghj Jun 30 '17 at 06:57
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Can you tell me how it came up in your mind that the base should be 3√2 ? – Fghj Jun 30 '17 at 07:20
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@RaviPrakash, $$\sqrt{18}=?$$ – lab bhattacharjee Jun 30 '17 at 07:22
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Because both sides can be written by $3\sqrt2$ to the power of something – Abdullah al allah Jun 30 '17 at 07:22
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Thank you to both. And without log too we can simply equate powers. – Fghj Jun 30 '17 at 07:23
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Let's try this way $$18^{4x-3}=3^{2(4x-3)}\times 2^{1(4x-3)}$$$$(54\sqrt2)^{3x-4}=3^{3(3x-4)}\times2^{\frac{3}{2}(3x-4)}$$and$$18^{4x-3}=(54\sqrt2)^{3x-4}$$so $$3^{2(4x-3)}\times 2^{1(4x-3)}=3^{3(3x-4)}\times2^{\frac{3}{2}(3x-4)}$$equating powers of $2$ or $3$ you get$$8x-6=9x-12$$$$x=6$$
Atul Mishra
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We have that $ ((3\sqrt{2})^2)^{4x-3}=((3\sqrt{2})^3)^{3x-4}$
So that means $(3\sqrt{2})^{8x-6}=(3\sqrt{2})^{9x-12}$.
Taking the logarithm of both sides with respect to base $3\sqrt{2}$ gives us $8x-6=9x-12$, which means $x=6$.
Saketh Malyala
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1We are able to rewrite the equation in terms of powers of $3\sqrt{2}$, so it logically follows to take the logarithm with respect to that base – Saketh Malyala Jun 30 '17 at 07:22
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1absolutely correct! i just chose to get rid of the powers @RaviPrakash – Saketh Malyala Jun 30 '17 at 07:24